Cl 


5iV^Bk.S  5 S°-E 


TRINITY  COLLEGE 
LIBRARY 

DURHAM,  NORTH  CAROLINA 


Rec’d 


Digitized  by  the  Internet  Archive 
in  2016  with  funding  from 
Duke  University  Libraries 


https://archive.org/details/elementsofalgebr01shou 


THE 


Elements  of  Algebra. 


BY 

F.  A.  SHOUP, 

PROFESSOR  OF  MATHEMATICS  IN  THE  UNIVERSITY  OF  THE  SOUTH;  LATE  PROFESSOR  OF  LOGIC, 
METAPHYSICS,  ETC.,  IN  THE  UNIVERSITY  OF  MISSISSIPPI. 


5 f l 8 


NEW  YORK: 

E.  J.  HALE  & SON,  PUBLISHERS, 


1874. 

IP 


Entered  according  to  Act  of  Congress,  in  the  year  1S74,  by 
E.  J.  HALE  & SON, 

In  the  Office  of  the  Librarian  of  Congress,  at  W ashington,  D.  C. 


Lange,  Little  & Co., 

PRINTERS,  ELECTROTVPERS  AND  STEREOTYPERS, 

108  to  114  Wooster  Street,  N.  Y. 


i '/  1 . o z. 
PREFACE.  S ; 


This  book  was  begun  with  the  hope  of  preparing  something  for 
beginners  in  Algebra  which  would  give  them,  in  a narrow  compass,  a 
philosophical  and,  therefore,  a thorough  start  in  their  analytical 
studies.  By  adopting  a somewhat  new  arrangement  of  the  subject, 
it  was  found  that  the  matter  the  author  had  in  mind  could  be  pre- 
sented in  such  a small  space,  that  he  determined  to  extend  the  scope 
of  the  work  far  enough  to  give  all  that  is  really  necessary  to  enable 
the  student  to  prosecute  with  profit  the  higher  mathematics.  This 
he  did  the  more  readily,  since  he  is  of  the  opinion  that  the  further 
course  in  Algebra  presents  rather  more  difficulties  than  any  other 
branch  of  mathematics,  and  that,  consecpiently,  the  student  at  the 
beginning  of  his  analytical  course  can  derive  little  real  benefit  from 
his  efforts  to  master  it.  In  truth,  so  far  as  his  information  goes,  it 
is  generally  either  omitted  altogether,  or  so  little  appreciated  by  the 
young  student,  as  to  be  almost  lost  labor.  It  is,  withal,  so  impor- 
tant, that  it  ought  to  be  insisted  upon,  and  so  should  be  taken  up  at 
a later  stage  of  the  student’s  course,  and  then  presented  in  connection 
with  the  general  philosophy  of  analytical  investigations.  Accordingly, 
if  this  little  work  meets  with  approval,  the  author  will  be  encouraged 
to  attempt  the  preparation  of  a supplementary  work  to  meet  this 
end. 

It  is  not  necessary  to  say  much  with  regard  to  this  essay.  It  will 
be  found  in  great  part  novel  in  its  treatment  of  the  subject.  The 
discussions  are  considerably  extended,  and  at  all  points  the  rationale 


4 


PEEFACE. 


of  operations  fully  brought  out.  An  effort  has  been  made  to  pre- 
serve the  continuity  of  the  subject,  so  as  to  present  a harmonious 
whole.  The  author  hopes  that  it  may,  in  some  measure,  prove  the 
means  of  stimulating  the  reasoning  powers  of  such  lads  as  may 
chance  to  use  it,  and  that  the  mere  mechanism  of  operations  will  not 
content  them. 


University  of  the  South, 
Sewanee,  Tenn.,  March , 1874. 


TABLE  OF  CONTENTS 


SECTION  I. 

ARTICLE 

Definitions  and  Explanations 1-17 

SECTION  II. 

ALGEBRAIC  TERMINOLOGY. 

Introductory  Remarks 18 

Translation  from  English  into  Algebraic  Language 10 

Translation  from  Algebraic  into  English  Language 20 

Translation  from  English  into  Algebraic  Language 21 

Translation  from  Algebraic  into  English  Language 22 

Translation  from  English  into  Algebraic  Language 23 

Numerical  Value 24 

SECTION  III. 

TREATMENT  OF  THE  + AND  — SIGNS. 

Introductory  Remarks 25 

The  Nature  of  the  Signs  + and  — 26 

Algebraic  Sum 27 

Addition  of  Quantities 28 

Subtraction  of  Quantities 29 

Subtraction  of  Polynomials 30 

Terms  Sum  and  Difference 31 

SECTION  IV. 

MONOMIALS,  EXPONENTS  AND  THE  SIGNS  X AND  -f-. 

Exponents  in  General 32 

Exponents  Entire  and  Positive. — Multiplication 33 

Law  of  Signs  in  Multiplication  and  Division 34 

Operations  on  Fractions 35 

Powers 36 

Roots • 37 

Division  of  Monomials 38 

Zero  Power 39 

Entire  and  Negative  Exponents 40 


6 


CONTENTS. 


ARTICLE 

Fractional  Exponents. — Radicals 41 

Fractional  Exponents. — Radicals 42 

To  Change  the  Index  of  a Radical 43 

To  Bring  Radicals  to  a Common  Index 44 

Product  of  the  nth  Roots 45 

Quotients  of  the  nth  Roots 45 

Division  of  Radicals 46 

Simplification  of  Radicals 47 

To  Pass  a Factor  under  the  Radical 48 

To  Transform  the  Sum  of  Radicals 49 

General  Principles  of  Exponents 50 

SECTION  V. 

TRANSFORMATION  OF  POLYNOMIALS. 

Introductory 51 

Multiplication  of  a Polynomial  by  a Monomial 52 

Multiplication  of  Polynomials 53 

Two  Terms  Unchanged  by  Simplifying 54 

Division  of  Polynomials 55 

Formulas  for  Transformations 56 

Formulas  for  Transformations 57 

Formulas  for  Transformations 58 

Factoring 59 

Factoring 50 

Division  of  the  Difference  of  Like  Powers 61 

Division  of  Like  Powers 02 

Binomial  Formula 03 

Powers  of  Polynomials 04 

Common  Multiple 05 

Operations  upon  Algebraic  Fractions 06 

Essential  Sign 07 

Imaginary  Quantities 08 

SECTION  VI. 

ROOTS  OF  NUMBERS. 

Extraction  of  the  Square  Root  of  Numbers 09 

Extraction  of  the  nth  Root  of  Numbers ‘0 

No  Exact  Root  of  an  Imperfect  Power 1 

SECTION  VII. 

EQUATIONS. 

Definitions 

Axioms N 

First  Transformation - 


CONTENTS. 


7 


ARTICLE 

Second  Transformation 75 

Third  Transformation 79 

Fourth  Transformation 77 

The  Change  of  Signs 78 

Solution  of  Simple  Equations 79 

Degree  of  Equations 80 

Complete  and  Incomplete  Equations 81 

Complete  Equations  of  the  Second  Degree 82 

Solution  of  Incomplete  Equations S3 

Solution  of  Complete  Equations 84 

Trinomial  Equations 85 

Simultaneous  Equations 86 

Elimination  by  Addition 87 

Elimination  by  Substitution 88 

Elimination  by  Comparison S9 

Elimination  in  general 90 

Equations  of  Condition 91 

Simultaneous  Equations  of  a Higher  Degree 92 

Simultaneous  Equations  of  the  First  and  Second  Degree  93 

Homogeneous  Equations  of  the  Second  Degree 94 

Solution  of  Simultaneous  Equations  of  any  Degree 95 

Radical  Equations 96 

Inequalities 97 

Problems...., 98 

SECTION  VIII. 

SYMBOLS  0 AND  00  . 

Meaning  of  the  Symbols  0 and  00 99 

Combinations  of  the  Symbols  0 and  co 100 

Vanishing  Fractions 101 

Problem  of  the  Couriers 102 

Problem  of  the  Lights 103 

Properties  of  Equations  of  the  Second  Degree 104 

Properties  of  Equations  of  the  Second  Degree 105 

Properties  of  Equations  of  the  Second  Degree 106 

Properties  of  Equations  of  the  Second  Degree 107 

Discussion  of  the  Four  Forms 108 

SECTION  IX. 

ARITHMETICAL  PROGRESSION". — RATIO  AND  PROPORTION. — GEOMET- 
RICAL PROGRESSION. 

Definition  of  a Series 109 

Arithmetical  Progression 110 

Formula  for  Last  Term Ill 

Sum  of  Means  at  Equal  Distances  from  Extremes 112 


8 


CONTENTS. 


ARTICLE 

Formula  for  Sum  of  Terms 113 

Eatio 114 

Proportion 115 

Product  of  the  Extremes 116 

Equi-multiples,  Powers,  Roots 117 

Inversion 118 

Composition 119 

Like  Parts 120 

To  Multiply  Proportions  together 121 

Sum  of  Antecedents  and  Consequents 122 

Continued  Proportion 123 

Mean  Proportional 124 

Reciprocally  Proportional 125 

Geometrical  Progression 126 

Formula  for  Last  Term 127 

Formula  for  Sum  of  Terms 128 

To  Find  Formula  for  any  Element 129 

To  Insert  Geometrical  Means • 130 

Sum  of  an  Infinite  Progression 131 

SECTION  X. 

LOGARITHMS. 

Definitions 132 

Characteristic 133 

Table 134 

General  Principles 135 

Solution  of  Equations 136 

Logarithms  of  Decimals 13 1 

General  Properties 138 

Modulus 133 

Indeterminate  and  Identical  Equations 1-40 

Indeterminate  Co-eificients 141 

Principle  of  Indeterminate  Co-efficients 142 

Developments 143 

Partial  Fractions 144 


The  Elements  oe  Algebra. 


SECTION  I. 

DEFINITIONS  AND  EXPLANATIONS. 

1.  Algebra  is  a branch  of  Mathematics  in  which  letters  are  com- 
monly used  to  represent  quantities,  while  the  operations  to  be  per- 
formed upon  them  are  merely  indicated. 

2.  The  leading  letters  of  the  alphabet  (a,  b,  c,  etc.)  are  used  to 
denote  quantities  whose  values  are  given,  or  may  be  assumed  at 
pleasure.  They  are  called  known  or  arbitrary  quantities. 

Any  numeral,  as  5 or  25,  is  made  up  of  a fixed  number  of  units,  and  admits 
of  no  change  whatever  in  this  regard  : 5 cannot  represent  25,  or  any  other  num- 
ber than  itself.  But  if  we  say  that  the  letter  a shall  stand  for  any  fixed  number 
whatever,  it  may  have  all  possible  values  at  pleasure.  It  cannot,  however, 
have  more  than  one  value  at  a time.  Its  several  values  must  be  taken  in  suc- 
cession. 

In  like  manner  a may  represent  any  kind  of  quantity,  as  pounds,  dollars, 
men,  etc.,  but  always  in  succession.  The  moment  we  attribute  a specific  nu- 
merical value  to  it,  it  ceases  to  be  an  algebraic  quantity  for  that  particular  value, 
and  becomes  an  arithmetical  quantity.  Whatever  is  here  said  of  a,  is  equally 
true  of  any  other  quantities,  b,  c,  d,  etc. 

3.  The  final  letters  of  the  alphabet  ( x , y,  z,  etc.)  are  used  to  denote 
quantities  to  be  determined.  They  are  called  unknown  quantities  ; 
or,  when  they  admit  of  an  indefinite  number  of  values,  in  succession, 
they  are  called  variables. 

Unknown  quantities  always  depend  for  their  values  upon  certain  other  quan- 
tities ; for  example,  if  we  say  x shall  be  a quantity  whose  third  part  shall 
always  be  equal  to  5,  the  value  of  x will  depend  upon  3 and  5,  and,  although 
unknown  for  a moment,  readily  becomes  known. 

If  we  should  take  two  quantities,  x and  y,  and  say  that  their  product  shall 
always  be  equal  to  a fixed  number,  as  100,  x may  be  5 and  y 20  ; or  x 4 and 
y 25,  etc.  Here  x and  y may  have  any  number  of  relative,  but  never  any  abso- 
lute, values.  In  such  a case  they  are  called  variables.  The  difference  between 
an  arbitrary  quantity  and  a variable  is,  that  all  arbitrary  quantities  may  be 
assumed  at  the  same  time  and  independently  of  each  other ; while  variables 
may  be  assumed  only  in  relation  to  each  other.  It  is  manifest  that  x and  y,  iu 


10 


ELEMENTS  OP  ALGEBRA. 


tlie  above  example,  could  not  both  be  assumed  at  the  same  time,  and  without 
regard  to  their  relative  values. 

4.  Since  quantities,  as  long  as  they  remain  purely  algebraic,  have 
no  fixed  numerical  value,  it  is  impossible  to  perform  any  arithmetical 
operation  upon  them ; such  as  addition,  multiplication,  etc.  All 
that  can  be  done,  therefore,  is  to  indicate  such  combinations  as  it 
may  be  desired  to  make.  For  this  purpose  certain  signs  are  em- 
ployed. The  sign  + indicates  addition  ; — , subtraction  ; x,  multi- 
plication y -r-,  division;  equality. 

If  a represents  the  number  of  days  one  man  works,  and  b the  number  another 
works,  since  a may  have  any  value,  and  b any  value,  we  manifestly  cannot  tell 
liow  great  their  sum  must  be ; but  we  can  write  them  so  as  to  show  that  their 
sum  must  be  taken,  if  their  values  ever  become  fixed  ; thus,  a + b (read,  a plus  b) 
indicates  that  these  quantities  must  be  added  together,  when  their  numerical 
values  are  determined  upon.  In  algebraic  language  they  are  said  to  be  already 
added  ; and  algebra  knows  nothing  of  any  other  kind  of  addition.  All  actual 
additions  must  be  made  by  the  laws  of  arithmetic.  The  same  may  be  said  of 
all  other  operations  upon  numbers. 

To  show  that  a quantity  is  to  be  subtracted  from  another,  we  write  the  quan- 
tity to  be  subtracted  after  that  from  which  it  is  to  be  taken  with  the  sign  — 
between  ; thus  a— b (read,  a minus  b)  shows  that  b is  to  be  taken  from  a,  or,  as 
is  commonly  said,  b is  subtracted  from  a. 

To  multiply  quantities  algebraically,  we  simply  connect  them  together  by  the 
sign  x ; thus,  axbxc  (read,  a multiplied  by  b,  multiplied  by  c)  shows  that 
their  product  is  required. 

It  is  far  more  usual,  however,  to  write  the  quantities  together,  thus,  abc,  with- 
out any  sign  between.  Sometimes  dots  are  used  ; thus,  a - b ■ c.  When  num- 
bers are  to  be  multiplied  together  the  sign  x must  be  used  ; thus,  4xo. 

To  show  that  one  quantity  is  to  be  divided  by  another,  we  may  write  them 
thus,  a-i-b  (read,  a divided  by  b).  It  is  much  more  common,  however,  to  write 

the  divisor  under  the  dividend,  thus,  j . 

b 

The  sign  = is  placed  between  quantities  to  show  that  they  are  equal  ; thus, 
a=b  (read,  a equal  to  b)  shows  that  an  equality  subsists  between  these  quanti- 
ties. The  quantity  to  the  left  of  the  sign  = is  called  the  First  Member;  that  to 
the  right,  the  Second  Member. 

5.  When  it  is  desired  to  show  that  two  or  more  quantities,  already 
connected  by  the  signs  plus  or  minus,  are  to  be  considered  as  a single 
quantity,  the  sign  ( ),  called  a parenthesis,  is  used;  thus,  (a  + b— c) 
shows  that  the  several  quantities  within  are  not  to  be  separated,  and 
are  to  be  considered  as  a single  quantity  so  long  as  the  parenthesis 
remains. 

This  sign  often  takes  the  form  [ ],  or  j j,  called  brackets.  A bar  over 


DEFINITIONS  AND  EXPLANATIONS. 


11 


several  quantities,  tlius,  a + b— c,  means  tlie  same  tiling1.  Also  tlie  quantities 
+ ax 

may  be  written  thus  — b ; here  the  quantities  on  the  left  of  the  bar  are  to  be 
+ c 

severally  multiplied  by  the  quantity,  x,  on  the  right. 

6.  The  sign  >,  indicates  inequality. 

The  opening  is  turned  towards  the  greater  quantity ; a>  b (read,  a greater 
than  b),  shows  that  a is  greater  than  b ; a <b,  shows  that  a is  less  than  b. 

7.  The  sign  oc,  shows  that  quantities  connected  by  it  vary  to- 
gether ; thus,  a oc  b means  that  a and  b increase  or  decrease  to- 
gether. 

s.  The  sign  is  an  abbreviation  for  therefore,  hence,  or  conse- 
quently ; v means  since  or  because. 

9.  An  Algebraic  Expression  is  any  quantity  or  combination  of 
quantities  written  in  algebraic  language. 

Thus,  | is  an  expression  for  the  quotient  of  any  two  quantities  ; ( a+b)c  is  an 
expression  for  the  product  of  the  sum  of  two  quantities  by  a third  quantity. 

10.  A Factor  is  any  quantity  which  enters  an  expression  as  a 
multiplier. 

Thus,  in  the  product  abe,  a,  b,  and  e are  each  factors  ; so  also  ab,  be,  and  ac 
are  factors  of  the  same  expression.  Unity  enters  every  expression  as  a factor. 
A factor  may  be  made  up  of  several  quantities  connected  by  the  sign  + or  — , 
thus,  in  ( a + b)c  and  (a+b)(a—b),  the  quantities  (a+b)  and  (a—b)  we  factors.  In 
fractional  expressions  unity  divided  by  the  denominator,  or  by  any  factor  of  the 

denominator,  is  a factor  of  the  expression;  thus,  in  - , is  a factor  of  in 

11.  A Co-efficient  (co-factor)  is  any  factor  of  an  expression. 

If  a numerical  factor  enters  an  expression,  that  factor  is  usually  written  first, 
and  is  especially  spoken  of  as  the  co-efficient.  Thus,  in  the  expressions  3 ab, 

fed,  and  — — ; 3,  §,  and  £ would  be  called  the  co-efficients,  respectively. 

Literal  factors,  however,  are  often  called  co-efficients  ; thus,  in  ax,  - • x , and 

b 

bey  ; a,  - , and  be,  would  be  called  co-efficients. 

So,  also,  numerical  and  literal  factors  taken  together  may  be  spoken  of  as 

co-efficients  ; thus,  in  the  expressions  3<z.r,  , and  3 (a  — b)z  ; 3 a,  — , and 

3(a— 6)  would  be  called  the  co-efficients  of  the  unknown  quantities,  respectively. 
Since  multiplication  is  but  abbreviated  addition,  each  factor  of  an  expression 


12 


ELEMENTS  OF  ALGEBRA. 


shows  how  many  times  all  the  other  factors  enter  it  additively ; thus,  in  the  ex- 
pression 3 ab,  the  3 shows  that  db  enters  the  quantity  three  times  by  addition,  so 
that  it  may  be  written  ab+ab+ab. 

So  if  we  had  a+a+a+a  we  could  evidently  write  4 a instead.  r + % + t 

0 0 0 

, na  3a  a a a a , , x 2a 

may  he  written  3r  or  — . r T - must  be  equal  to  — . 

b b b b b b b 

If  we  had  (a+b)+(a+b)+ written  c times,  we  could  write  c(a+b). 

Here  c or  ( a+b ) may,  either  of  them,  he  regarded  as  the  co-efficient. 


12.  An  expression  composed  altogether  of  simple  factors  is  called  a 
monomial. 

By  simple  factors  is  meant  those  which  are  composed  without  the  aid  of  the 
signs  + or  — . 

ab,  — , 3ab,  are  monomials.  A monomial  is  spoken  of  as  a term,  especially 
c 

when  found  connected  with  other  quantities  hy  the  sign  plus  or  minus  ; thus, 

in  the  expression  7 ax——^+g,  any  one  of  the  three  monomials  which  enter  it  is 

a term.  Sometimes  complex  expressions  tied  together  hy  the  parenthesis,  or 

otherwise,  are  called  terms;  thus,  in  a—(b+c)  and  — , the  expres- 

’ c b—c 

sions  (a  + c),  and  may  he  called  terms. 


13.  A Binomial  is  an  expression  composed  of  two  terms  ; a Trino- 


mial has  three  terms.  a + b, 


dab— 3c,  ^ 


4 b 
ac ' 


are  binomials. 


cl 

ac  + fi—S, 


and 


3d 


4«  — 5c  + , are  trinomials 


A Polynomial  is 


an  expression  composed  of  two  or  more  terms;  thus,  a + b,  5 ab 


2 

+ ~ 
a 


- + etc.,  are  polynomials. 


d,  a + b + c 

14.  When  the  same  factor  enters  an  expression  more  than  once,  as 
in  3 aaciabbccc,  the  expression  can  be  greatly  shortened  by  writing 
any  such  factor  but  once,  with  a small  figure  to  the  right,  and  a little 
above,  to  show  the  number  of  times  the  quantity  enters  as  a factor; 
thus  the  above  expression  would  take  the  form,  3 aib~cz,  read,  three  a 
to  the  fourth  power,  b to  the  second  power,  c to  the  th  ird  power.  The 
small  figure  so  used  is  called  au  Exponent  or  Index,  because  it  shows 
the  number  of  times  the  quantity,  to  which  it  is  affixed,  enters  as  a 


factor. 

It  is  important  to  distinguish  clearly  between  exponents  and  num- 
bers which  enter  an  expression  as  factors,  called  co-efficients.  In  the 
above  example,  the  3 written  first  is  a multiplier,  whereas  the  small 


DEFINITIONS  AND  EXPLANATIONS. 


13 


figures  do  not  enter  the  expression  ns  quantities  at  all ; they  are 
merely  signs,  to  show  how  many  times  other  quantities,  namely  a,  b, 
and  c,  enter  as  factors. 

An  exponent,  therefore,  is  any  quantity  used  to  show  how  many 
times  another  quantity  must  be  taken  as  a factor. 

Letters  are  often  used  as  exponents;  thus,  a,  5°,  {a  + bf,  a/,  6",  (read,  a 
to  the  x power,  etc.,  a to  the  one-half  power,  b to  the  m divided  by  n power.)  Any 
symbol  of  quantity  may  be  used  for  the  same  purpose. 

15.  Terms  are  said  to  be  Homogeneous,  when  they  contain  the  same 
number  of  literal  factors;  thus,  9 a2bz,  2 cxi,  25x3y2,  abcxij,  are  all 
homogeneous.  They  are  said  to  be  homogeneous  with  respect  to  a 
certain  quantity  or  class  of  quantities,  when  there  are  the  same  num- 
ber of  such  quantities  in  each  ; thus,  ax2,  b2cxy,  Hiy2,  are  homogene- 
ous with  respect  to  the  unknown  quantities  which  enter  them. 

A polynomial  or  an  equation  is  homogeneous  when  all  of  its  terms 
are  homogeneous. 

16.  The  Reciprocal  of  any  quantity  is  unity  divided  by  that  quan- 


Thus,  -,  — r,  , are  the  reciprocals  of  a,  a—b,  and  a2b. 

17.  The  Square  Root  of  a quantity  is  indicated  by  placing  over  it 
the  sign  ; when  any  other  root  is  to  be  taken,  a small  number  is 
written  to  the  left  and  above  the  sign,  to  show  the  degree  of  the  root 
required;  thus,  ^/,  indicates  the  third  root ; ty,  the  fourth  root ; 
%/,  the  with  root,  and  the  sign  is  used  thus,  Va,  *\/a2b3,  Va2b  + d, 


. The  small  figure  used  to  show  the  degree  of  the  root  re- 


quired, is  commonly  called  the  index. 

This  sign  grew  out  of  the  custom  of  the  older  algebraists  of  writ- 
ing r,  signifying  root,  before  the  quantity  whose  root  was  required. 
When  used  before  an  expression  tied  together  by  a bar,  thus, 
r • a + b,  it  would  take  very  nearly  the  form  of  the  present  radical 
sign. 


tity. 


Ill 


14 


ELEMENTS  OF  ALGEBRA. 


SECTION  H. 


ALGEBRAIC  TERMINOLOGY. 

18.  Algebra  has  a language  of  its  own,  which  must  be  thoroughly 
mastered  before  the  subject  can  be  at  all  comprehended.  In  the  fore- 
going Section  we  have,  so  to  speak,  learned  the  alphabet.  Yfe  may 
now  proceed  to  use  it.  Let  it  be  borne  well  in  mind  that  it  is  all  pure 
conventionality.  Any  other  signs  or  symbols  might  have  been  used; 
but  now  that  we  have  agreed  to  use  those  already  explained,  we  must 
adhere  to  them  strictly. 


19.  First  let  us  translate  from  English  into  Algebraic  symbols. 

1.  Add  together  any  tivo  quantities. 

Explanation. — Since  they  must  be  any  two  quantities,  we  cannot  take  partic- 
ular numbers  ; and  since  numbers  from  their  nature  are  always  definite,  we 
cannot  take  numbers  at  all.  a and  b are  two  such  quantities,  and  since  their 
values  must  remain  undetermined,  we  can  only  write,  a + b.  Any  other  letters 
would  have  done  as  well. 


2.  Write  the  difference  of  any  two  quantities.  Ans.  a — b. 

3.  Write  the  product  of  any  two  quantities.  Ans.  ab. 

4.  Write  the  quotient  of  any  two  quantities.  Ans. 

5.  Write  the  product  of  the  sum  and  difference  of  two  quantities. 

Ans.  ( a + b ) ( a—b ). 

G.  Write  the  quotient  of  the  sum  and  difference.  Ans. 


Remark. — When  a quantity  has  2 for  an  exponent  it  is  often  read  square,  and 
though,  perhaps  not  strictly  correct,  it  is  convenient  to  retain  the  custom  ; 
thus  a2  may  be  read  a square.  When  a polynomial  has  2 for  an  exponent,  as 
(a  + b)2,  it  must  be  read  a plus  b squarED.  In  like  manner  a3  may  be  read  a 
cube;  (a  + b)  3,  (a  + b)  cub ed. 


7.  Write  the  sum  of  the  squares  of  two  quantities. 

8.  The  square  of  the  sum. 

9.  The  square  of  the  difference. 

10.  The  square  of  the  quotient. 


Ans.  a2  + b2. 
Ans.  (a  + b)2. 
Ans.  (a—b)2. 


11.  The  sum  of  the  square  roots. 

12.  The  square  root  of  the  sum. 


Ans.  \Za  + Vb. 
Ans.  y'a  + b. 


13.  The  product  of  the  square  roots. 

Ans.  \>ra  x or  \/ a.  Vb. 


ALGEBRAIC  TERMINOLOGY^ 

14.  The  square  root  of  the  product. 

15.  The  quotient  of  the  square  roots. 

16.  The  square  root  of  the  quotient. 

17.  The  square  root  of  the  product  of  the  squares. 

Ans.  V a2  b2. 

18.  The  square  root  of  the  product  of  the  sum  and  difference. 

Ans.  V(a  + b)  ( a—b .) 

19.  The  cube  root  of  the  product  of  the  cubes.  Ans . V a3  b 3. 


15 

Ans.  AJab. 

A A fa 

Ans.  — — . 

Vb 

Ans. 


20.  Translate  the  following  expressions  into  English : 
1.  a—b,  a + b,  a2+b2,  ab,  a2b2,  \. 


2. 

3. 

4. 

5. 


a2—b2,  (a—b)2,  a2+b2,  (a  + b)2,  (a  + b)  (a—b). 
a2  a + b a—b  a+b  (a  + b)2  a — b 
b2'  a ’ b ’ a—b’’  a2 +b’  (a—b)2’ 


Va  + Vb,  's/a  , Vb,  Vab, 

Vb 


V a—b,  V a2  +b2  • 


V (a  + b)(a—b). 


6. 


V7 a2+b 2,  Va2// 

a 2/~a/ 

T ’ y lA’ 


V(a  + b)2,  Va2-b2,  Va2-b2, 


21.  Translate  the  following  into  algebraic  language  : 

1.  The  square  of  the  sum  of  two  quantities  is  equal  to  the  square 

of  the  first,  plus  twice  the  product  of  the  first  by  the  second,  plus  the 
square  of  the  second.  Ans.  (a  + b)2  = a2  + 2ab  + b2. 

2.  The  square  of  the  difference  of  two  quantities  is  equal  to  the 

squai'e  of  the  first,  minus  twice  the  product  of  the  first  by  the  second, 
plus  the  square  of  the  second.  Ans.  (a—b)2=a2—2ab  + b2. 

3.  The  product  of  the  sum  and  difference  of  two  quantities  is 
equal  to  the  difference  of  their  squares.  Ans.  (a  + b) (a—b)  = a2  — b2 . 


16 


ELEMENTS  OF  ALGEBRA. 


4.  The  quotient  of  the  difference  of  the  squares  of  two  quantities 
by  the  difference  of  the  quantities,  is  equal  to  the  sum  of  the  quanti- 


ties. 


Ans. 


«2  — b2 


— Oj  -f  b . 


5.  The  quotient  of  the  difference  of  the  squares  of  two  quantities 
by  the  sum  of  the  quantities  is  equal  to  the  difference  of  the 
quantities.  a2  — b2 

a +b 


- — a—b 


G.  The  product  of  the  sum  of  two  quantities  by  the  first  is  equal 
to  the  square  of  the  first,  plus  the  product  of  the  first  by  the  second. 

Ans.  ( a + b ) a = a2  + ab. 

Let  the  algebraic  expressions  in  this  article  be  translated  back  into 
English. 


22.  Translate  the  following  expressions  into  English : 

1.  \fa  x Vb  = V ab. 

Ans.  The  product  of  the  square  roots  of  two  quantities  is  equal  to  the  square 
root  of  the  product. 


3.  (Va+  Vb){Va  — Vb)  = a — b . 

4.  (Va  + Vb)~  = a + 2Vab  +b. 

5.  {Vet— Vb)~  = a — 2V~ctb  + b. 

6.  V ) Va=a3,  Va~  ~ 

l » -i 

L y a — an,  Vam  = a”>  V a~m  — — . 

nm  * 


9.  («  + 3)£  = V a Ab,  [(a  + 5)2]i  = a + b. 

10.  (ai+bi)(ai—bi)  = a—b. 


23.  Express  in  algebraic  symbols  the  following  : 

1.  Divide  a certain  quantity  into  three  equal  parts. 

a a a 

o 5 q J Q * 

o o o 


ALGEBRAIC  TERMINOLOGY.  17 

2.  Divide  a given  quantity  into  two  parts,  one  of  winch  shall  be 

three  times  the  other.  , a 3 a 

Jins.  — . 

4 4 

3.  What  two  numbers  are  those  which  differ  from  each  other  by  a ? 

Ans.  x and  x—a;  or,  x and  x + a. 

4.  The  a part  of  a quantity  added  to  its  b part  is  equal  to  m. 

Ans.  — + r = m. 
a b 

5.  Three  times  a certain  number  minus  that  number  is  equal  to 

f the  number  less  a.  Ans.  3x—x  — f- x—a. 

6.  The  sum  of  two  numbers  is  to  the  difference  of  those  numbers 

as  m is  to  n.  Ans.  a + b : a — b ::  m : n. 

7.  The  ratio  of  the  square  of  the  sum  of  two  numbers  is  to  the 
sum  of  the  squares  of  those  numbers  as  m is  to  n. 

Ans. 

(i a + b)2  m 


24.  If  at  any  time  particular  values  are  given  to  the  quantities  in 
an  expression,  the  indicated  operations  may  then  be  performed.  A 
result  so  found  is  called  the  numerical  value  of  the  expression.  For 
each  new  set  of  values  for  the  several  quantities  which  enter  an  ex- 
pression, there  will  be  a new  result  as  the  numerical  value  of  such 
expression. 

The  numerical  value  of  the  expression  ab  — ~ , when  a = 4,  and  b = 2,  gives 

4 x 2 — 1 = 6,  which  is  the  value  of  the  expression  in  this  case. 

If  a = 7,  and  6 = 5,  we  should  have  7 x 5 — \ = i§-a.  , 


Find  the  numerical  values  of  the  following,  letting  a = 1,  b = 2, 
c — 3,  and  d — 4 : 


1. 


cd  ab 

3a>-j’j  + 


a—  b 


(a  + b)2  2(Z 
cd  a * 


2.  (a  + b)(a  + b),  *t*+*^±,  [( a + b)(c-a)]d . 

\/ b + a Vd-C  abed  V c — b 

3.  Vd  [rt(5+c)]V«,  x ^Ta  ’ VbIIaX  abcd  ' 


55—3  ^ 1 (a+b—2)(c  — a + d) 

4 6* — a b ’ c\5  d)\c  a)  / a c\  la  a\ 

\b+d)  \b~~~c) 


B 


18 


ELEMENTS  OP  ALGEBRA. 


SECTION  III. 

TREATMENT  OF  THE  + AND  - SIGNS. 

25.  All  elementary  operations  of  Algebra  fall  under  one  of  the 
two  following  heads,  namely : 

I.  Indications. 

II.  Transformations. 

As  we  have  seen  in  the  previous  section,  Indications  are  the  opera- 
tions of  expressing  in  algebraic  language  whatever  may  be  stated  in 
spoken  language. 

Transformations  comprise  all  changes  which  may  be  lawfully  per- 
formed upon  expressions  after  they  are  written  in  algebraic  symbols. 

The  simplest  transformations  are  those  made  upon  quantities 
connected  by  the  signs  phis  or  minus. 

Thus,  a + a + a-\-a,  mav  evidentlv  be  written  4 a.  This  is  a change 
of  form  without  a change  in  the  numerical  value. 

a + a—a  + a = 2a,\s  another  such  change. 

Algebraists  have  commonly  called  these  transformations  addition; 
but  manifestly  the  algebraic  sum  is  altogether  accomplished  in  the 
operation  of  Indication.  The  quantities  are  no  more  added  after 
the  form  is  changed  than  when  they  are  simply  connected  together 
by  their  proper  signs. 

26.  The  Nature  of  the  Signs  + and  — . 

Let  us  now  try  to  understand  the  nature  of  the  signs  + and  — . 

Their  first  and  most  obvious  use  is  to  connect  quantities  together 
so  as  to  show  that  they  are  to  be  added  or  subtracted.  But,  then, 
certain  quantities  may,  from  their  nature,  be  additive,  while  others 
are  subtractive,  before  any  combination  takes  place.  For  example, 
a man,  casting  up  his  accounts,  would  consider  all  amounts  due  him 
to  be  augmentative,  while  all  amounts  which  he  owed  would  be 
diminutive  of  his  capital.  The  first  class  would  take  the  sign  + ; 
the  second  the  sign  — . It  is,  however,  a mere  matter  of  agreement 
as  to  which  shall  be  called  plus  and  which  minus  ; hut  they  are  al- 
ways contrary  the  one  to  the  other,  so  that  the  establishment  of 
' either  determines  the  other.  Quantities  affected  with  the  sign  + are 
called  Positive  ; those  with  the  sign  — are  called  Negative. 

To  illustrate  this,  suppose  we  take  a right  line,  ab,  and  agree  to 
reckon  all  distances 


a 


o 


b 


TREATMENT  OF  THE  + AND  — SIGNS. 


l'J 


to  the  right  of  the  point  o as  positive ; then,  of  necessity,  those  to 
the  left  would  be  negative.  If  we  had  determined  to  reckon  dis- 
tances to  the  left  of  o positive,  those  to  the  right  would  have  been 
negative,  and  so  in  general. 

The  sign  of  a quantity  must  be  known  before  we  can  use  it  in  com- 
bination with  other  quantities.  Plus  is  always  understood  where  no 
sign  is  written.  A quantity  with  the  signs  + and  — , thus  ±a,  is 
to  be  used  first  positively  and  then  negatively. 


27.  To  find  the  algebraic  sum  of  several  quantities. 

When  quantities  are  given  with  their  respective  signs,  to  find  their 
aggregate,  or,  as  it  is  called,  their  algebraic  sum,  we  have  this  simple 
principle : 

Write  the  several  quantities  one  after  the  other  in  any  order,  con- 
nected by  their  respective  signs. 

Examples. 


1.  Add,  dab,  4 a2b,  2 ab,  —ab,  and  —5 a2b. 

A ns.  3ab  + 4jrb  + 2ab—ab—5a2b. 

J __  Q I 

2.  Add,  —la,  —j- , —be,  cAtP,  and  'fa. 


Ans.  — — j—  — 1 
d 


%a—bc  + a ~b\  + V a. 


Remark. — It  is  usual  to  place  a positive  quantity  at  the  beginning. 

3.  Add,  + 24,  -\fWUi,  -iy,  and  (a  + b)(a-b). 

Ans.  + 24— \/67a  + j^- + («  + £)(«— &). 

, , , , a 2a  5a  , , /a\m  , /a\m+n 

4 Add,  — V ’andU)  ' 

, 2a  a 5a  , , /a\m  [ a\m+n 

Am-  TS- J + l + i+{b)  + (s) 

5.  Add,  7 a2—  2 aW,  3«2  + 4 ab-,  and  2 a—am+  5 aT 

Ans.  7a2—  2ab2  + 3 a2  + 4 crfF  + 2 a—am+  5 oh 


28.  To  reduce  a polynomial  to  the  least  number  of  terms. 

Terms  are  said  to  be  like  or  similar  when  they  contain  the  same 
letters,  and  the  several  letters  have  the  same  exponents,  respectively. 
2 x^bmc3  and  5aAbm<?  are  like  terms.  The  algebraic  sum  of  several 


20 


ELEMENTS  OF  ALGEBBA. 


quantities  can  often  be  much  shortened  by  gathering  into  one  all 
the  terms  which  are  alike. 

For  example,  3«#  + 4«2J  + 2aZ>— ab— 5a2b  may  be  written  4«Z> — 
a2b  ; for,  3 ab  and  2 ab,  both  being  +,  give  5 ab;  but  — ab  (that  is,  one 
ab  to  be  taken  away)  leaves  4 ab',  so  4 a~b  less  5a2 b,  leaves  one  a2b 
minus. 

So  in  genera],  to  reduce  a polynomial  to  the  smallest  number  of 
terms  we  may  say: 

Gather  like  terms  together,  giving  the  results  their  proper  signs, 
respectively. 

Great  care  must  be  taken  to  combine  only  terms  whose  literal  parts  are  en- 
tirely the  same,  and  to  add  and  subtract  the  numerical  parts  according  to  their 
signs. 

Examples. 


Reduce  the  following  polynomials  to  the  smallest  number  of 
terms : 

1.  2«— 4«2J2 +3«  + 5a252— 4m  Ans.  a + a-b 2. 

2.  3b%c— 2bic  + am— bm+'+5am— 2bm+\  Ans.  bic  + Gam—3bm+\ 

n o,  cd  ncd  „ . , _ 

3.  ■=  + g + 2-  - 3—  + g + 24  -12. 

b a J b a 

. 3 a j:d 

Ans.  — 2 1-  12. 

b a 

4.  3 a2 + 5 ambn  — a' 2 d + Gai b + 2a' 2 d - 2 ambn — aib. 

A ns.  4 a2  c * + 3ambn  + oaib. 

5.  a2  +2ab-\-b2  +a2 — 2ab  + b2  = 2a2  + 2b2. 

6.  a + 2\/ab  + b + a — 2\Zab  + b—2a  + 2b. 

7.  V«  + 2 \/rab-{-  Vb+  V«— 2\fab  + \Gb=2\fa  + 2*Jb. 

8.  2(a+5)i-3(a-S)i  + 5(a+5)i-2(ffl-5)i 

Ans.  7(a  + &)^— 5(a— b)i. 


9 ^a  + h 

~7c 


/a— cV  A /a  + b /a— cV  am~x 

\bd)  7/T  + \bd)  ' b-1' 


V c 


. 2 VaA-b  a1”-1 

Ans.  1-  y— r. 

V c b 

10.  25a2b*  c + 12amb’1  — 5a2b3c  + amb-— 3a2b3c  + 5amb’. 

Ans.  17«2Z>3c+18«’”Z>3. 


TREATMENT  OP  THE  + AND  — SIGNS. 


21 


29.  Subtraction  of  quantities. 

To  subtract  a positive  quantity,  as  + b from  a,  we  should,  strictly, 
write  a — ( + £),  using  the  parenthesis  to  prevent  the  confusion  of 
signs;  but  since  + b is  but  another  way  of  writing  b itself,  we  may 
drop  the  + sign,  and  the  parenthesis  with  it,  and  write  a — b.  When 
"the  quantity  to  be  subtracted  is  already  negative,  as  —b,  the  indica- 
tion is  made  in  the  same  way,  a— (—6);  but  we  cannot,  as  before, 
drop  the  — sign  of  the  quantity,  since,  in  that  ease,  we  should  have 
for  a result,  a—b  as  before,  and  should  thus  have  taken  away,  not 
—b,  but  + b.  To  take  away  a negative  quantity  is  really  to  augment 
the  quantity  from  which  it  is  said  to  be  taken  ; thus,  (25—5)  is  20; 
now  from  (25—5)  remove  the  —5,  that  is,  strike  it  out,  and,  of  course, 
the  25  will  be  left.  By  taking  away  —5  we  have  really  added  5 to 
the  quantity  (25—5),  from  which  it  was  taken;  thus,  25  — 5 — (—5) 
=25—5  + 5=25,  or  in  general,  a — b — {—b)=a  — b + b=a. 

A man  who  is  insolvent  owes,  say,  a dollars.  It  would  be  the  same 
thing  to  say,  he  has  —a  dollars  due  him.  To  subtract  or  take  away 
any  part  of  his  debts,  is  really  to  augment  his  fortune  by  just  that 
amount ; and  so  in  general, 

To  subtract  a negative  quantity,  is  to  aclcl  the  numerical  value  of 
the  quantity  ; or  practically, 

To  subtract  any  quantity,  write  it  after  the  quantity  from  which 
it  is  to  be  taken  with  its  sign  changed. 


Examples. 


1.  From  a take  b,  ab,  —cl,  3b,  —2ab,  +cl,  and  then  simplify  the 
resulting  expression. 

Ans.  a—b—ab  + cl—3b  + 2ab—cl  = a—Ab  + ab. 


2.  From  ci  + b take  b,  —c2,  3c,  —2b,  and  simplify. 

A ns.  a + b—b  + c2—3c  + 2b  — a + 2b  + c2—3c. 

3.  From  3*^  —cd  take  ^ —cel  and  . 

b b o 


A ns.  3 T-cd- 
o 


a 7 am  _ a a” 
'l+ci~W~2i~¥ 


4.  From  a + yfab  + b take  a — 'fob  and  —b. 

A ns.  a A V ab  + b—a+  V ab  + b = 2 *f~ab  + 2b. 

11  A 1 1 1 

5.  From  a 2 — b?  take  2ci2,  —3 b2,  —5 a2,  and  b2. 

A ns.  a?  — b?—2a2  -f-3bi  + 5ai  — b~=-la^  +bi. 


22 


ELEMENTS  OF  ALGEBRA. 


6.  Prom  2 V a + 3Vb  take  — Va,  2 Vb,  i\ fa,  and  — Vb. 

A ns.  2 Va  + 3V b -\-  V a — 2 Vb  — i Va  + Vb — | V a + 2 Vb. 

7.  From  take  -3-  ^ 4 SZ+Q.  and  -J-  a~b 


Am.  ^ + 3-  “ 

c c c 


_i  a~b 
*'  c -*“■ 


8.  From  V ab— c take  — 2V~ab,  5 c,  ^ ^ and  —6c. 

M«s.  V ab— c + 2\/ab  — 5c— ^-^  + 6c  = \V~ab- 

9.  From  v^— take  9y/a,  — 2aVb,  and  — Sy«. 

^4«s.  -\/a— 2aV~b— 9Vff  + 2-v/£  + 8v/ft  = 0. 


30.  Subtraction  of  polynomials. 

To  indicate  the  subtraction  of  any  polynomial,  we  have  but  to 
write  such  polynomial  within  a parenthesis,  and  connect  it  by  the 
— sign  with  the  quantity  from  which  it  is  to  be  taken  ; thus,  to  take 
a + b—c  from  cl,  we  have  cl  — («  + b— c). 

Now  the  minus  sign  here  means  that  we  are  first  to  get  the  alge- 
braic sum  of  all  the  quantities  within  the  parenthesis,  and  then  take 
this  sum  from  d.  Manifestly  the  same  thing  can  be  accomplished  by 
taking,  first  a,  and  then  b,  and  then  — c,  away  from  cl;  thus,  cl— a— 
b + c.  The  parenthesis  has  disappeared,  and  the  several  quantities 
have  changed  their  signs.  We  may  then  say,  that, 

To  subtract  a polynomial  from  any  quantity,  we  have  but  to  write 
the  severed  terms  in  succession  after  the  quantity  from  which  it  is  to 
be  taken  with  their  signs  changed. 


Examples. 

1.  From  3a  — 2b  take  a+Aib  and  simplify. 

Ans.  3a — 2b — a—Tb=2a — 6b. 

2.  From  an— a take  am  + an— a2. 

Ans.  an—a—am—an  + a-=a-—am—a. 

m m 

3.  From  a»5-5- (-1  take  — 3««£-2  + l. 

Ans.  a » b~*  + 1 + 3anb~'J—  1 = du f b~\ 


TREATMENT  OF  THE  + AND  — SIGNS. 


23 


4.  From  8 ap-V  take  oof  lcq  + a » c'!+-p_1  — j/«n. 

1 cq oap  \f—a  *c?fp  =. 5ap~1cq — a~ficq+7l~1 + ^/~a^. 

5.  Simplify  a~—bm—(2a'~—bm— 1). 

Ans.  a~  — bm— 2a^  + bm  + l = l—dfi. 

6.  Simplify  a2  — 7a2b3c— J—  (a2 — 7ci2b3c  + i). 

Ans.  a2  — 7a2b3c—\—a2  + 7 a2b3c—\—  —1. 

7.  Simplify  x2  — x— 1 — (— x2  + .T  + 1). 

Ans.  x2  —x—l  + x2  —x—  1 =2x2  —2x—2. 

Remark. — It  will  be  observed  that  in  removing  a parenthesis  from  a poly- 
nomial when  it  has  the  minus  sign  before  it,  we  change  the  signs  of  all  the 
terms  within.  When  the  sign  before  the  parenthesis  is  plus,  the  parenthesis 
may  be  removed  without  any  other  change. 

Eemove  the  parentheses  and  simplify  the  following: 

8.  a—(—b  + c—a)=a  + b—c  + a=2a  + b—c. 

9.  l + (a—b—l)=l+a—b—l=a—b. 

10.  a — (—ct)=2a;  a — (—l)  = a + l. 

11.  3 a2b-  ~ -(-3 a2b+~]=6a2b~  j. 

12.  or  — 1 + ( — cir  + 1)  = 0. 

13.  -*  + l-(-*  + l + V3)  = -V& 

14.  am~l—bff—(am-'  — b~k  + l)  = — 1. 

15.  1— %a+(l— 1-«)=2— a. 

16.  a—(—a)  — (—a)  — (2a)=a. 

Remark. — We  may  put  a parenthesis  upon  the  algebraic  sum  of  several  quan- 
tities without  any  other  change,  provided  that  the  positive  sign,  stands  before  the 
parenthesis  ; but  if  the  negative  sign  occupies  that  place,  the  signs  of  all  the 
terms  within  must  he  changed. 

Enclose  the  following  in  parentheses,  first  with  a + sign,  and 
then  with  a — sign. 

1.  a— b + c=(a— b + c) 

= —(—a  + b—c). 


24 


ELEMENTS  OF  ALGEBRA. 


3.  a=(a)  ; 1— rt=(l— «);  am—  ( am ) 

= — (—1),  l—a=  — ( — l + a),  = — (— am). 

Remark. — Where  there  are  several  parentheses  or  brackets,  one  within  another, 
in  removing  them  begin  by  removing  the  outer  one,  and  then  the  next,  and  so 
on.  We  may,  however,  begin  with  the  inner  one. 


Kemove  the  brackets  from  the  following : 

1.  — [a— {a  + b — 1)]  = — a + {a  + b— 1)=£— 1. 

2.  a—  [ — { — {a  + 5)}]  —a  + { — (a  + b)} —a— (a  + b) =a—a—b=  — b 

3.  1 — [«—(«  + b)  + {a—b)+b\  — 1 —a  + (a  + b)  — (a—b)  — b = l— a 
+ « + b— a + b— b=l—  a + b. 


The  management  of  fractional  quantities  in  Algebra  and  Arith- 
metic is  entirely  the  same. 

Now,  since  tve  can  separate  a fraction,  such  as  ]Ut°  as  many 
partial  fractions  as  tve  please  to  break  the  numerator  into  parts,  thus 
^ + 1 y + t5t,  we  may  do  the  same  thing  with  an  algebraic  fraction. 


Now,  if  such  a fraction  as  this,  having  the  algebraic  sum  of  two 
or  more  quantities  for  its  numerator,  has  the  — sign  before  it,  and  it 
should  be  thus  broken  into  parts,  the  removal  of  the  division  bar 
will  act  altogether  like  the  removal  of  a parenthesis.  The  place  of 
the  bar  must,  thei’efore,  be  supplied  by  a parenthesis ; or  the  signs 
of  the  partial  fractions  must  all  be  changed. 


Examples. 


4.  l-{  + [-(-l)]}=0. 

5.  -(!  + (— 1))  = — 2. 


For  example,  — , may  oe  written,  ^ ^ ^ 


Examples. 


Separate  the  following  into  partial  fractions  : 


a b c 

-7  — i 


1. 


d 


TREATMENT  OF  THE  + AND  — SIGNS. 


25 


2. 


a + b—c 

I 


_a  b 
~d+d~ 


c 

d' 


2a2b—cm  _ /2 a2b  c”  \_  2a25  cm 

3a"c2  \3anc2  3anc2)  3anc2  ^ 3anc2 ' 


4. 


—ai+b 
a + b 


i 

aJ 


a + b 


+ 


b 

a + b ’ 


Remark. — Tlie  sign  — standing  before  the  division  bar  shows  that  the  whole 
fraction  is  to  be  subtracted  ; or,  in  other  words,  that,  after  all  the  operations 
indicated  are  performed,  the  result  is  to  be  subtracted ; but  when  the  sign 
stands  before  the  first  term  of  the  numerator,  as  in  the  last  example,  it  affects 
only  that  term. 


5. 


— 5 + 4a— 2? 


-5_  4a_  _2+ 

+ 5“  5” ' 


6. 


—5  + 4a— 23 
5“ 


(_5_  4a  _22\_  5 
\ 5"  + 5m  5mJ~5‘ 


4 a 2^ 

5"  + 5" 


7. 


a + b 
a—b 


c\_  fa  b 
d)~  \a—b  a—b 


c\_  a b c 
d)  a—b  a—b  d‘ 


8. 


a + b 
(a  + b)2 


a—b 

(a  + b)m- J — 

r a b / a b \1_ 

L (a  + b)2  + (a  + b)2  \(a  + b)m  (a  + b)m) 

a b a b 

(a  + b)2  (a  + b)2  + (a  + b)m  (a  + b)m’ 


31.  Meaning-  of  the  terms  SUM  and  DIFFERENCE. 

The  terms  Sum  and  Difference  must  be  understood  in  Algebra  in 
their  largest  sense.  To  add  does  not  necessarily  mean  to  augment, 
nor  does  subtraction  always  mean  a diminution;  thus,  —b  added  to 
+ b gives  0 ; while  —b  subtracted  from  +b  gives  2b. 

By  the  sum  we  are  to  understand  the  result  obtained  from  con- 
necting the  quantities  by  their  own  signs  ; a difference  results  when 
certain  terms  are  connected  with  their  signs  changed. 


26 


ELEMENTS  OF  ALGEBRA. 


SECTION  IV. 

MONOMIALS— EXPONENTS  AND  THE  SIGNS  x AND  h-. 

32.  An  Exponent  is,  by  definition,  a sign  which  shows  how  many 
times  the  quantity  to  which  it  is  affixed  is  to  he  reckoned  as  a 
factor. 

Exponents  may  be  entire  or  fractional,  positive  or  negative. 


33.  Entire  and  Positive  Exponents. 

53  is  the  same  as  5 x 5 x 5,  or  125  : a3  is  but  another  way  of  writing 
a • a • «;  a2b3  — a-a-b-b-b;  (a  + b)2  = (a  + b)(a  + b) ; {\fa)3  = 
V a ■ ''fa  • fcr,  am—a  • a • a • a written  m times. 

Quantities  united  by  the  sign  x or  — are  simply  made  co-factors; 
thus,  3 a2b  x5ab 3 shows  that  the  continued  product  of  all  the  factors 
is  required.  The  sign  x may,  however,  be  dropped  altogether  and 
the  quantities  be  written  together,  where  no  confusion  is  like  to  re- 
sult; thus,  3a2boab3 ; but  in  such  a case  as  this  it  is  usual  to  retain 
the  x or  use  the  • 

Now  the  order  in  which  factors  are  taken  makes  no  difference  in 
the  product;  so  that  we  may  write  the  above  expression  thus, 
3 x 5 a2abb3.  It  will  be  observed  that  the  sign  x must  now  be  used 
between  the  numerals  to  prevent  mistake. 

But  here  we  have  the  indicated  multiplication  of  two  factors  which 
can  be  actually  performed ; so  that  15  can  be  written  in  the  place  of 
3x5. 

We  see,  further,  that  a enters  three  times  as  a factor,  so  that  a 2 a 
may  be  written  a3.  In  like  manner  bb3  is  equal  to  b 4.  It  is  thus 
manifest  that  3 a2b  x 5ab3  = l5a3bi. 

We  have  simply  multiplied  the  numerical  factors  together  and 
written  each  letter  once  with  an  exponent  equal  to  the  sum  of  the 
several  exponents  of  that  letter.  Transformations  of  this  kind  are 
called  multiplication. 

\<le  may  say,  then,  in  general,  that : 

To  multiply  monomials  together,  multiply  the  numerical  factors 
together,  and  after  this  product  write  the  several  letters  ivliich  enter 
the  monomials,  giving  to  each  an  exponent  equal  to  the  sum  of  all  the 
exponents  of  that  letter  in  the  several  terms. 


MONOMIALS — EXPONENTS  AND  THE  SIGNS  X AND  -P. 


27 


Examples. 

1.  Multiply  4 a2bc  by  5 ale2. 

2.  Multiply  a:d2f  by  2 atl3f2. 

3.  Multiply  xmy  by  xmy. 

4.  Multiply  xm~ly  by  xym+\ 

5.  Multiply  5 axm  by  3 anxn. 

6.  Multiply  25 arlfc  by  3 a3bac. 

7.  Multiply  l~a2b3  by  \ab2. 

8.  Multiply  | axyn  by  f avyz*. 

9.  Multiply  ambncp  by  abc. 

10.  Multiply  %am+nb  by  -§ am+nb. 

11.  Multiply  xm+nbn+l  by  lzxm~nb]~n. 

12.  Multiply  xm~n+lbr~q  by  xn~m~'ibq~p . 


Ans.  20 a3b2c3 
Ans.  2«°c/s/3 
Ans.  x^y2 
Ans.  xmym+ 
Ans.  15  an+'ixn+n 
Ans.  75ap+3ba+2c!! 

Ans.  %a3b5 
Ans.  %ax+vyn+lz2 
Ans.  am+1br,+icF+l 
Ans.  -i ^aim+Snb3 
Ans.  \xlmV1' 
Ans.  x°b° 


Remark. — We  shall  find  that  the  same  laws  govern  the  management  of 
fractional  and  negative  exponents  as  when  they  are  positive  and  entire,  so  that 
we  may  employ  them  in  our  examples  at  once.  In  finding  the  sum  of  fractional 
exponents,  let  the  student  remember  to  apply  the  rules  for  the  addition  or  sub- 
traction of  fractions  in  arithmetic. 


, _ l 1 

13.  a-i  x a# = a. 

14.  abc  x aibzd=aibici. 

15.  ixzy~3  x \xmy  a —\:C^y. 

11  5 3 

16.  5x3y2z  x axy~azn  = 5axiy'Zz2. 

17.  ar-bc%  xa3b~id=ab~1c. 

m p m p 

18.  25  amb^  x 5«"&?==125am+MZ>»  ?. 

19.  ambncp  x crmb~nc=a°b°cp+ '. 

nn  aAbcx~l  aFbac  aA+p  ba+ccx~x 

20.  — - — x— = yw 


34.  The  law  of  signs  in  Multiplication  and  Division. 

So  far  no  mention  has  been  made  of  tlie  signs  of  the  terms  multi- 


28 


ELEMENTS  OP  ALGEBRA. 


plied  together.  Let  us  now  inquire  what  effect  these  signs  will  have 
upon  the  product. 

In  the  multiplication  of  two  quantities  wTe  simply  repeat  one  of 
them  additively  as  many  times  as  there  are  units  in  the  other.  Let 
us,  then,  multiply  two  quantities,  a and  b,  together,  and  first  let  them 
both  be  positive.  + a must  be  added  to  itself  until  there  are  alto- 
gether as  many  times  a as  there  are  units  in  b,  so  that  we  shall  have 
+ ab. 

Thus  we  see  that, 

The  •product  of  two  positive  quantities  is  positive. 

Now,  let  a be  negative  and  b be  positive.  — a must  now  be  added 
to  — a until  — a has  been  repeated  b times.  But  — a and  — a gives 
— 2a;  —a  and  — a and  — a gives  — 3 a;  and  when  taken  b times 
we  must  have  — ba. 

If  a had  been  positive  and  b negative,  — b would  have  been  taken 
additively  a times,  and  the  result  would  have  been  again  — ab.  Thus 
we  see  that, 

The  product  of  a positive  and  negative  quantity  is  always 
negative. 

Now  let  us  take  a and  b both  negative.  We  shall  have  (—  a ) 
(—  b).  But  (—a)  = — (+«);  and  so  we  may  write  — (+  a)  (—  b) 
= — (—  ab)  — + ab.  Thus  we  see  that, 

The  product  of  a negative  quantity  by  a negative  quantity  is 
positive. 

Since  the  product  of  the  divisor  and  quotient  must  always  produce 
the  dividend,  the  same  principles  hold  good  in  division. 

The  laAV  of  signs  may  then  be  summed  up  as  follows: 

In  multiplication  and  division  like  signs  give  plus,  and  unlike 
minus. 


Examples. 


1.  Multiply  5 a~b  by  — 

2.  Multiply  — l-ambn  by  amb. 

3.  Multiply  — \ambn  by  — amb. 

4.  Multiply  \ambn  by  — 5 amb. 

5.  Multiply  Urb"  by  ■§ amb. 

6.  Multiply  —9 am  + lb~2  by  5 abmc. 


5 

Ans.  — 35rtV&3. 


A ns.  — ia°mb"  + l. 
Ans.  $a*"bn  + 1. 


Ans.  — | a-”b'  H. 
Ans.  \armbn  + x. 


Ans.  — 45«”'  + 5&m  ~c. 


MONOMIALS — EXPONENTS  AND  THE  SIGNS  X AND  = . 


29 


IX-  _1  —1 - 

7.  Multiply  a&bsc*  by  — 3a  ~b  3c  ». 


Ans.  — 3 a°b°c°. 


8.  Multiply— a°b~lc  by— 1. 

9.  Multiply  ab°c~l  by  a~"bc. 

10.  Multiply  ab°c~x  by  ab~°c. 


Ans.  a°b~'c. 
Atis.  a}~"bc°. 


Ans.  a 2. 


Remark. — When  a quantity  has  an  exponent  zero,  such  exponent  shows  that 
the  quantity  does  not  enter  as  a factor  at  all.  It  may  therefore  be  omitted  alto- 
gether. Remember,  however,  that  the  quantity  itself  is  not  zero.  Any  quantity 
to  the  zero  power  is  thus  1 ; but  we  shall  see  further  upon  this  point. 

11.  abxabx  — ab  = —azbz. 

i i Li 

12.  —ax%  x —ax 3 x ax-  = azx  e . 


It  has  already  been  said  that  an  algebraic  fraction  does  not  differ 
from  a numerical  fraction  in  principle. 

In  the  treatment,  therefore,  of  algebraic  fractions,  we  have  but  to 
apply  the  laws  of  arithmetic  to  those  already  established  for  the 
management  of  algebraic  symbols. 


15.  10 am-nb~p  x — a°  = -10 am~nb~p. 


16.  «(— «)(—«)«(— a)  = — «5. 

17.  ( — 1)(— 1)(-1)(— «°)  = 1. 

18.  (— 0(-  6°)(—  o°)  = - 1- 

is.  (hmri})°v~a  = vs. 


35.  Operation  upon  Fractions. 


Examples. 


1.  Multiply  | by  |. 

2.  Multiply  — | by  ~. 

3.  Multiply  -|by  -C-. 

4.  Multiply  — | by  ^ . 


30 


ELEMENTS  OP  ALGEBRA. 


5.  Multiply  — — bv  — . 

— b " c 

6.  Multiply  - p by  . 

7.  Multiply^  by  ^ . 

^\rr  u ^ 7/P 

8.  Multiply  by  -1. 

_ « — « « — a «4 

9.  T X -7—  X r X t = 7— 


abc 

— 1 

10. 

X -7—  = 1. 

■^1 

abc 

an 

lrm 

11. 

y-  X 

— x — 2—- 

bm 

a 

a° 

b a 

12, 

— X 

77  X 7 = 1, 

a 

b°  b 

-A 

—be 


a 


3 

2 


,4  MS. 


cCV 


—a2b3 
A ns. 


a2b ' 
15  2?* 

w 


A ns. 


OX”  'yp 

Sab-'  ' 


36.  The  Powers  of  Quantities. 

The  Poiver  of  a quantity  is  the  product  obtained  by  using  the 
quantity  a certain  number  of  times  as  a factor ; thus,  the  second 
power  of  3 is  9,  the  third  power  of  2 is  8.  a 2 is  the  second  power  of 
a ; a3,  the  third  power ; am,  the  mth  power  of  a. 

To  form  the  poiver  of  any  quantity  multiply  the  quantity  by  itself 
as  many  times  as  there  are  units  in  the  exponent  of  the  poiver  less 
one. 

A practical  rule  for  finding  the  power  of  a monomial  is : 

Raise  the  numerical  factor,  if  there  be  one,  to  the  required  poiver 
and  multiply  the  exponent  of  each  letter  by  the  exponent  of  the  power. 

This  process  is  called  Involution. 


Examples. 

1.  (a2)2,  (2a)5,  (—3 z&)2,  (-£.^)3,  (-cr1)5,  (am)n,  {a»Y. 

3 Pm 

Ans . a4.  32a5,  9x,  — — a“5,  a*”,  a » . 


MONOMIALS — EXPONENTS  AND  THE  SIGNS  X AND  -L. 


31 


Remark. — It  will  be  observed  that  tlie  sign  of  an  ecen  power  must  always  he 
positive  ; and 

Tliat  the  sign  of  an  odd  power  must  always  be  the  same  as  the  sign  of  the 
term  itself. 


Ans. 


144  aW 
49a3 


iri  y3 


Cl  ® 


V*  ’ 

bT 


a 


a 


a » 


Ans.  1,  v— tjtt-j- — • — , j 
’ b-i’b2m’  A\ 

a n bz  b 


ar  aA  2 7a~3b3 


4a4  b 


37.  The  Boots  of  Quantities. 

The  operation  of  extracting  the  root  of  a monomial  is  just  the  re- 
verse of  that  of  raising  it  to  a power.  Practically, 

Extract  the  required  root  of  the  numerical  factor,  if  there  beany, 
and  divide  the  exponents  of  the  titered  factors  by  the  index  of  the  root. 

This  process  is  called  Evolution. 

If  a quantity  be  raised  to  an  even  power,  the  sign  of  the  power 
must  be  plus,  whether  the  quantity  (that  is  the  root)  is  positive  or 
negative;  thus, 

a3— a x a or  —ax  —a, 

a4  =a  x ax  ax  a or  —a  x —a  x —a  x —a. 

Whence  it  follows,  that  when  the  index  of  the  required  root  is 
even,  the  root  may  have  either  the  plus  or  minus  sign ; it  is  there- 
fore given  both ; thus, 


32 


ELEMENTS  OF  ALGEBRA. 


Hence  we  may  say  that, 

The  even  root  of  a quantity  has  the  double  sign  ±. 

When  the  root  is  odd,  the  sign  of  the  power  must  always  he  the 
same  as  that  of  the  root;  thus, 

—ax—ax—a=—a3;  ( — a)5  = —a5;  (— a)2"^1  — — a2m+r 

axaxaxa  — +a3 ; { + a)5—+a&-,  ( + «)2"1+1  = +a2m+1 

It  follows,  therefore,  that, 

An  odd  root  always  has  the  same  sign  as  that  of  the  quantity 
itself. 


Examples. 

Extract  the  roots  of  the  following,  as  indicated: 

1.  VaS  Via2,  V^a2,  V9a462>  VSa3,  V — 8a3,  V\a6b3. 

Ans.  ±a2,  ± 2a,  ± \a,  ±3a2b,  2 a,  —2a,  l-a2b. 

Ans-  V”  ±8“‘s"’  a’a>3- 


d C a3c1d,  fiabdh,  f 36«  ^ A 


^ 3 jj  1112  rn  0^ 

Ans . , 2 ,namb‘mcm,  ±6a  ~b  2,  — j. 

])n 

4.  V — 27a3,‘V 

/u\i 

Hns.  —3anl2,  3amb 2, y->  ’ 

5.  VT,  ^ • 


-I  "I  "1  ^ ^ 

21^5.  i 1?  1?  1?  t)  ? 4)  * 


There  is  no  such  thing,  so  far  as  the  human  understanding  can 
reach,  as  the  even  root  of  a negative  quantity;  for  we  cannot  conceive 
of  a quantity  which  is  not  either  + or  — : but  the  product  of  any 


MONOMIALS — EXPONENTS  AND  THE  SIGNS  X AND 


33 


quantity,  + or  — , taken  an  even  number  of  times  as  a factor,  is  +; 
therefore,  since  an  even  root  must  enter  the  power  an  even  num- 
ber of  times,  we  cannot  conceive  of  such  a negative  power;  thus, 
V— a2  can  be  neither  + a nor  — a ; for  -fax  + a=+a2  and  — ax 
—a=+a2  : so  that  — a2  cannot  be  produced  by  the  multiplication 
of  any  thing  whatever  by  itself.  Such  indicated  roots  are  called 
Imaginary.  In  general, 

An  imaginary  quantity  is  the  indicated  even-  root  of  a negative 
quantity ; thus, 

V — 4,  a/— a,  V— 3,  a3b2,  and  V — 

are  all  imaginary. 


38.  The  Division  of  Monomials. 


Division  and  Multiplication  are  reciprocal  operations.  Any  quan- 
tity united  to  another  by  the  sign  x may  be  replaced  by  its  recipro- 
cal (unity  divided  by  it)  with  the  signs-;  thus,  5x2  = 5 ---J-  or 

a x l— a -j-  ^ . Whatever  has  been  said,  therefore,  of  multiplication, 

taken  in  the  converse  sense,  will  apply  to  division. 


The  signs  x and  always  indicate  operations  upon  factors.  The 
sign  x means  to  add  as  a factor ; and  means  to  subtract  as  a fac- 
tor. a3  xa  shows  that  the  quantity  a3  is  to  be  further  augmented 
by  the  introduction  of  another  a as  a factor,  giving  a4 ; while 
«3-i-  a shows  that  a factor  a is  to  be  withdrawn  from  a3,  giving  a2. 

To  divide  one  quantity  by  another,  is  to  withdraw  the  divisor  as  a 
factor  from  the  dividend.  Where  a factor  of  the  divisor,  or  the  di- 
visor itself,  enters  the  dividend  exactly,  it  maybe  taken  out  at  once; 
thus. 


15 a2bicd 
3ab3c 


= bahd. 


Numerals  are  disposed  of  as  in  arithmetic,  and  where  the  same 
letter  is  found  in  dividend  and  divisor,  the  difference  of  their  expo- 
nents gives  the  new  exponent  of  that  letter  in  the  quotient. 

Let  it  be  remembered  that  like  signs  give  4-  and  unlike  — . 


Examples. 

1.  Simplify  2 §<Tb(?d  = 5 o?c. 

2.  Simplify  9 w’lAdx*  -f-  3a3cx. 

c 


A ns.  baled. 
Ans.  aa'lf'cx. 


34 


ELEMENTS  OF  ALGEBRA. 


3.  Simplify 

4.  Simplify 

5.  Simplify 


am  + xbhf 
-aTbhj  ' 

— asb*xn 
t ibx 2 

—ambnyr 
-aby  ’ 


Ans.  — aby. 
A ns.  —a?bzn~-. 
Ans.  an~1bn~lyp~ 


39.  The  Zero  Power  of  a Quantity. 


Iii  dividing  one  quantity  by  another  it  often  happens  that  the  ex- 
ponents of  the  same  letter  in  the  two  quantities  are  the  same,  in  which 

case  that  letter  disappears  entirely ; thus,  (~  = a. 


We  may,  however,  leave  the  letter  in  the  quotient  by  giving  it  the 

exponent  zero  ; thus,  "^-—ab°.  Here,  manifestly,  b°=  1.  The  zero 

simply  shows  that  b is  not  a factor  of  the  quotient  at  all.  To  com- 
prehend how  any  quantity  to  the  zero  power  must  be  equal  to  unity, 
we  have  but  to  remember  that  unity  enters  every  expression  as  a 
factor,  and  that  thus  a°  = 1 .a°. 

Now,  removing  a,  since  the  exponent  ° shows  that  it  is  not  a 
factor,  the  co-efficient  1 remains. 


1. 


Examples. 

oa^bci  25  ambic,  — xAifq 


ci2b 


-5  a'"b 


1 —a 
xy  ’ — 1 ’ a 

1 _ L “ _ , r _j 

Ans.  3c-,  — 5 b * c , — x*  y*  , 1,  — 1. 


©V1) 


0 a°  —a0  3 ax°  \x°{y°) 

' ~Za  ’ -(4)°  ’ (25)°  • 

Ans.  1,  — 1,  1,  4,  — 1. 

(5 )mxny3  ( a + by  (a  — b)m(a  + b)n  — [n  + («s64);] 


( o)nx  ly2’  ( a + b)'! ’ ( a—b)a  ' — [ + ( — !)] 


Ans.  bm~nx*  + y,  (a  + b), 


(. a-b)m-\a  + by 


■,  —a—a*bs. 


4.  — = am~n=a°.  But  = 1 a°  = 1. 


MONOMIALS — EXPONENTS  AND  THE  SIGNS  X AND  -r- . 


35 


40.  Entire  and  Negative  Exponents. 

When  the  exponent  of  a factor  in  the  divisor  is  greater  than  the 
exponent  of  the  same  letter  in  the  dividend,  the  subtraction  of  the 
greater  from  the  less  must  give  a negative  exponent;  thus, 


3«25 


= 3 a-'b. 


Now,  in  this  case,  a enters  the  numerator  but  twice  as  a factor, 
so  that  when  it  becomes  necessary  to  take  out  four  such  factors,  we 
shall  have  a deficiency  of  two.  This  fact  the  — sign  shows  : that 
is  to  say,  we  must  still  withdraw  the  factor  a twice. 

This  meaning  of  the  negative  exponent  follows  from  the  general 
definition.  A negative  exponent,  then,  shows  that  not  only  is  there 
no  factor  such  as  that  over  which  it  is  written  in  the  quantity,  but 
that  such  quantity  must  still  be  diminished  by  that  factor  a certain 
number  of  times. 


or1  =- : 
a' 


for 


ai~s=cr\ 


But 

a 6 a 


-i  1 

, a = -. 
a 


In  general,  let  p and  q be  two  quantities  whose  difference  is  m,  q 
being  the  greater ; then, 

ap 


But  - = — . 

fi7  am 


= — . That  is, 
cC  ’ 


Any  quantity  raised  to  a negative  power  is  equal  to  unity  divided 
by  the  quantity  raised  to  the  corresponding  positive  poioer. 


Again, 

cf_  1 _ 1 . , , 

d1  «2_s  a-1  ’ U 


az  1 

— = a:,  a — — = 


a;  a 

whose  difference  is  m,  q being  the  greater. 

cf  _ JL_  _ JL 
ap  ~~  a p_7  ~ a~ 


or,  in  general,  let  p and  q be  two  quantities 


but 


a 7 
a f 


— a" 


a = 


That  is, 


36 


ELEMENTS  OF  ALGEBRA. 


Any  quantity  raised  to  a positive  power  is  equal  to  unity  divided 
by  that  quantity  raised  to  a corresponding  negative  poiver. 

It  follows  from  the  foregoing  that, 

Any  factor  may  be  transferred  from  the  numerator  to  the  denom- 
inator ; or  from  the  denominator  to  the  numerator  by  changing  the 
sign  of  its  exponent. 


Examples. 

Convert  the  following  expressions  into  fractional  forms  with  unity 
for  numerators : 

1 1 m 

1.  a,  ab,  a s,  am,  a~b3,  5 ab»,  20 z~yy~. 

1 1 1 I 1 1 1 

a a b a a~nb~3  o-'a-'b—  ^ y 


2.  ( a + b)~\  \ , a~'bm,  x » if . —a,  —(—a)  K 


Ans. 


(a + o)’  (a\  ctb  m’  a 


1 -1  -1  _ 1 
’ ~-1’  —a  a' 


Convert  the  following  into  expressions  containing  no  negative 
exponents. 

„ 0 _!  a~mb  o-'xy~?  (a  + b)-1  (a—b)(a  + b)-1 

6 • 2-'ab~m’  ( a-b )->’  {a-b)-\a+b)‘ 

8 b 2 2 xbm  a—b  (a—b)(a—b) 

AnS‘  a’  tfV  (a  + J)(a+J)# 

Convert  numerators  into  denominators,  and  the  converse. 


4. 


a~^bm  ^ahcx~3  9 ~'xy 
2c~ld * cd~p  ’ — (ZV-’  -1 

2-,cf?-1  d~mc~q  (-I)”1 


Ans. 


7~'a  'b'  ah~m  (f ’ 9‘r  ^ ' 


Remark. — It  must  be  carefully  borne  iu  mind  that  only  factors  can  be  tlius 
transferred.  Thus,  in  the  expression  tlie  a or  6 canuot  *>e  written  in 

the  denominator.  The  a + b must  be  taken  together,  thus,  ^ 7 


MONOMIALS — EXPONENTS  AND  THE  SIGNS  X AND  -4-. 


37 


We  could  write,  however, 


1 


or 


c 


c 


1 1 


a a 


41.  Fractional  Exponents.-  Radicals. 

From  the  general  definition  of  an  exponent,  it  follows  that  a frac- 
tional exponent  must  show  how  many  times  the  quantity  to  which 
it  is  affixed  is  to  be  reckoned  as  a factor : that  is  to  say,  that  the 
quantity  is  to  be  resolved  into  as  many  equal  factors  as  there  are 
units  in  the  denominator  of  the  exponent,  and  that  as  many  of  these 
factors  are  to  be  taken,  as  there  are  units  in  the  numerator ; thus 


Now,  the  Root  of  a quantity  is  that  factor  of  it  which,  taken  a 
certain  number  of  times,  will  produce  the  quantity  itself. 

If  there  are  but  two  equal  factors,  either  of  them  is  the  square 
root ; if  there  are  three,  any  one  of  them  is  the  cube  root ; if  there 
are  m equal  factors,  any  one  of  them  is  the  mth  root ; thus, 

25—5  x 5 .*.  5 = V25;  27=3x3x3  .-.3  = ^27.  In  like  man- 

ner, if  a contains  b m times  as  a factor,  b—’\/a. 

It  follows  therefore,  that, 

The  denominator  of  a fractional  exponent  shows  the  root  to  be  ex- 
tracted ; thus, 


am  = y a. 


38 


ELEMENTS  OF  ALGEBRA. 


It  is  thus  manifest,  that, 

The  radical  sign  is  identical  in  signification  with  a fractional  ex- 
ponent, having  unity  for  its  numerator  and  the  index  of  the  radical 
for  its  denominator.  Any  radical  may,  therefore,  be  removed  by  en- 
closing the  quantity  in  a parenthesis,  and  writing  instead  of  it  such 
a fractional  exponent. 

There  is  thus  no  necessity  for  the  use  of  the  radical,  and  it  is  only 
retained  here  because  it  is  so  universally  found  in  mathematical 
works. 


Examples. 

Transform  the  following  into  expressions  with  fractional  expo- 
nents : 


1.  a/5,  V/c,  ■y/d,  Va  + b. 


1 1 l_  J_ 

Ans.  52,  c3,  d«,  ( a + b )-. 


Remark. — When  the  radical  sign  is  removed  from  a polynomial,  as  in  the 
last  example,  the  bar  must  be  retained,  thus,  a + b- , or  replaced  by  a parenthesis, 


thus,  (a  + &)2 


; so,  also,  where  any  mistake  might  result,  as  in  or  y , the 


parenthesis  must  be  used  ; thus,  (§)2  and  J. 


/25xm  _ (25)  - (xm) 2 3 / a-b~  _ a3b'c 

3'V  *r'~  sV')*’ 


b)3' 


vV“  = 


V ]/ 


(O’1 


- = X ri . 


5.  Vax  Vb=a^b^=  (ab)^,  \Za-Vb=anb<* . 

6.  j/ Va  = ( y/a^j  - = (V2 ) 2 = aT  , |/ V n — ( V7^) ”*=  («" ) " = ■ 

7.  a~i=  \/a , a3!)*  — ^/~aA  • y/b3  , cfrbdc'1  =Va./y/b(V\t 


MONOMIALS — EXPONENTS  AND  THE  SIGNS  X AND  -P. 


39 


43.  Fractional  Exponents.— Radicals. 

Now,  while  the  denominator  of  a fractional  exponent  shows 
the  degree  of  the  root  to  be  extracted,  the  numerator  shows  the 
number  of  times  such  root  is  to  be  taken;  that  is,  it  shows  the  power 
to  which  this  root  is  to  be  raised ; thus, 


S3=22=4. 


But,  (v/a)”=  -y/cT,  for  we  may  write  (f/a)m=an  = ia”')n  = \/am\ 
hence,  we  can  say  in  general,  that 

The  numerator  of  a fractional  exponent  shows  the  power  to  which 
the  quantity  affected  by  it  is  to  be  raised,  while  the  denominator  slioius 
the  root  to  be  extracted. 


Examples. 


Transform  the  following  into  equivalent  expressions  : 

f «"*,  yV'?,  ’vV 


m, P _»  lm  5?  1 

1.  v an,  ai , a*,  a2  , an,  ««+». 


n 9 U / 1 

Ans.  a™,  V" /if  — , 

2.  j/ ffa2,  ^a2‘Vaz‘l\/ai,\/ 

Ans.  (^)-==  («•)",  a^aTav,  (^)K^) " =-^Vr- 


3.  a/28- V~5- v'a. 


An,  SW-*  = & 


4. 


a*Jb  5 \/ a2  3 a2b(cd)™  cd{  25)! 
cV~d4  5 6ffa3 


4 ax 

ab ¥ 5 a»  3 a2bycd  bed 
Ans.  j , 3" . ^ 

cd~  6«4  4 


’ 4 a.r' 


40 


ELEMENTS  OF  ALGEBRA. 


43.  To  change  the  Index  of  a Radical. 

It  is  obvious  that  we  cau  multiply  or  divide  the  numerator  and 
denominator  of  a fractional  exponent  by  the  same  thing,  without 

2 772  2m  yni  m 

changing  its  signification ; thus,  a2—aA,  oA—om,  aT^—a ».  But  the 
numerator  is  the  exponent  of  the  power  and  the  denominator  is  the 
index  of  the  root;  whence  we  may  say,  in  general,  that 

We  can  multiply  or  divide  the  index  of  a radical  by  any  quantity, 
provided  that  ioe,at  the  same  time,  multiply  or  divide  the  exponent  of 
the  quantity  under  the  radical  sign. 

For  example,  take  \/a2  +b—c.  If  we  should  multiply  the  index 
of  the  root  by  2,  we  must  raise  a2  +b— c to  the  second  power;  thus, 
vV2  +b— c = V' (a2  +b—c)2.  The  quantity  under  the  radical  must 
be  considered  as  one  quantity.  We  may,  however,  multiply  or  divide 
the  exponent  of  each  factor  under  the  sign;  thus,  a/3«26  = 
i\/9aib2,  or  \/‘66aib6=  \/6a2b3. 


Examples. 


• Change  the  indices  of  the  following  by  multiplying  by  2. 

1.  *fa,  Vo,  ‘s/x,  \/‘da2b,  f/?>a2b,  \/‘ia"‘b". 

Ans.  V^  v^2,  VWtfb2,  a/9 a^b2,  f/ 9 

Multiply  by  3. 


\/a  \/a  + b Vl5 a2x~1  i/ a 2 V a—b 

‘ V ’ ~ v'j  ’ * r 

ty{a  + b)2  v/(15)3«6.t-3  ,{/  a~^ 

W’  * ’ ^3  ’ v ’ 


Ans. 


\/  (a—b)3 
\/  (a  — b)3 


Multiply  by  m. 


MONOMIALS — EXPONENTS  AND  THE  SIGNS  X AND 


41 


Divide  by  2. 

4.  y/tf,  V64a:*r4, 

^4  ms.  V a,  \/%azb,  8 xy~s, 


6 a~1b'h 
5:% 2 


44.  To  bring1  Radicals  to  a common  Index. 

Radical  quantities  with  different  indices  may  be  changed  into 
equivalent  radicals  having  the  same  indices.  For  let  'sfaP  and  \/bq 
be  two  such  quantities.  They  may  be  written, 

p J? 

ffm  and  bn. 

Causing  these  fractional  exponents  to  have  a common  denominator, 
we  have, 

pn  qm 

amn  and  bmn  ; 

and  thus,  ’^/aT  and  "f/W.  We  may,  therefore,  say  that, 

To  bring  radicals  to  a common  index,  find  a common  multiple  of 
the  several  indices,  and  take  this  for  the  common  index ; then  divide 
this  neiv  index  by  each  of  the  old  ones  in  succession,  and  raise  each 
quantity  under  the  radicals,  respectively,  to  the  power  indicated  by 
these  quotients;  or 

Use  f ractional  exponents  and  bring  them  to  a common  f ractional 
unit. 

For  example,  take  -|-v/3 b2  and  a\/c3.  The  least  common 

multiple  of  the  indices  is  12.  We  may  then  write  4 V"  ",  iV  , 
aV  ; now  raising  the  quantities  under  the  radicals,  respectively,  to 
the  powers  found  by  dividing  12  by  2,  by  3 and  by  4,  we  have, 

4v^(2a)#,  4^(3 b2)*  and  aV(c3)3;  or 
4v/,64 a6,  Vv' 81 58  and  a'\/ c9. 

It  will  be  observed  that  the  quantities  outside  of  the  radicals  re- 
main unchanged. 

Examples. 

Transform  the  following  radicals  into  equivalent  ones  having 
common  indices: 


42 


ELEMENTS  OF  ALGEBRA. 


1.  l-Voa2, 


y%bc, 


Ans.  l-\/l 25«6, 


VWb2, 


2.  ao/b,  ^v/3«, 


A 2 3.  A 

4.  a2,  &3.  A«s.  «6,  6b. 

5.  (a  + 6)2,  («-5)t 


Am.  f/ 

7/i  p qvn  pn 

a* , bT.  Ans.  as* , h* . 
Ans.  ( a + iy , (a—b)c. 


45.  Product  of  the  «th  Roots. 


The  product  of  the  nth  roots  of  two  quantities  is  equal  to  the  nth 
root  of  the  product  of  the  quantities,  and  the  converse. 

For, 

i i i 

cin bn  = {ah) n ; whence, 
f/ a • f/b  = f/ ah. 

Again, 


The  quotient  of  the  nth  roots  is  equal  to  the  nth  root  of  the  quo- 
tient, and  the  converse, 

For, 

i i 


whence, 


yA 

f/h 


These  principles  enable  us  to  combine  radicals  by  multiplication 
or  division.  Thus, 


a'fT  x c\/7i=acV~b  • ‘\/d=-ac\/'hd ; 

hence, 

To  multiply  radical  quantities  together,  multiply  the  factors  out- 
side of  the  radical  signs  together,  and  also  the  factors  under  the  radi- 
cal signs,  retaining  the  radical  over  the  latter  product. 

For  example, 

3 a2h\f  hcdm  x 2ab3  VAd—Sa^b  x 2ab3  V ocdm  x cd=6a3bi\/bc3d '"~1 . 


MONOMIALS — EXPONENTS  AND  THE  SIGNS  X AND  -4-. 


43 


The  indices  of  radical  expressions  so  combined  must  be  the  same ; 
if  not  so  already,  they  must  be  changed  into  corresponding  radicals 
with  a common  index. 


Examples. 


1.  3a\fb*  xab\f7. 

2.  3a(b2)?  x5b(7)?. 

3.  %Vax\Vbx 

4.  x x - 


3 a2b  3 r-7-  irr 

5.  v 4c  x \ V 


X 

3 a2b 


(4c)‘^  x UiY*. 


7.  — V'2a  x \/2a  x 4 Y/2a. 

8.  Y/ an  x — v'g”  x — Yfa. 

n m 1 

9.  a™  x —an  x —a7. 

10-  zj/Y^f/fr 


11.  2(~V  x — c(-^ 


12.  5aV —3x  x —2b\/2x2. 

13.  5a(-3x)i  x -2b(2x2y*. 


Ans.  15  abV”b2. 
Ans.  15ab(7b2)2. 

Ans.  —\Vb. 
Ans.  — -jt(J)2. 


Ans. 


3 aH 
2x 


j/f- 


- my- 


Ans.  -iy8192a13. 
Ans.  m\/ an-p+m'2p+mn. 

n2p  m2p  mn 

Ans.  amnp^~mnpJr mnp' 


Ans. 


. / ac 

icym 


Ans.  —2c 


Ans.  — lOab  V — 108a;7. 


Ans.  — 10«5(— 27«3)6(4a;4)'3  — — 10«5(  — lOS-r1)6, 


46.  Division  of  Radicals. 

To  show  how  to  divide  one  radical  expression  by  another,  we  have, 

aYfb  a Y/h  a ,m/b 

-s-=  =-•—?=—  -\/  ^5  whence, 

cYjd  c Y/d  c\  d 

To  divide  one  radical  quantity  by  another , divide  the  quantities 


44 


ELEMENTS  OF  ALGEBRA. 


■without  the  signs,  and  the  quantities  within,  respectively,  retaining 
the  radical  sign  over  the  latter  quotient. 

The  indices  must  be  made  common,  if  not  so  already. 


Examples. 


3Va  _ 3 , /a 


3(a)* 

2# 


3/a\i 

2W  ’ 


3u2(k)~ 


5. 


3.r(2 a*Y  _ 3g/210g>0\^y 


2yV 


2 7 J 


8«25(5«52)» 

2a6(7c6~1)^ 


MONOMIALS — EXPONENTS  AND  THE  SIGNS  X AND  -f-. 


45 


47.  Simplification  of  Radicals. 

Any  factor  may  be  removed  from  within  a parenthesis  and  made  a 
factor  without,  by  multiplying  its  exponent  by  the  exponent  of  the 
parenthesis;  thus, 

2 a(b*cMdy=  2ab4(c3mdy. 

2a(lfcZmd)2  = 2 ab(c3md)^. 

2a(b2c3md)i  = 2«cm(5VZ)x. 

1 11 

2 a(b‘2c3mcl)m  = 2abnc3d"‘. 

Numerical  factors  may  be  treated  in  the  same  way ; thus, 
2a(12b*c3d)s =2a(2*  x 3b*c*cd)i 
—2  x 2abc(3cd)^ 

=iabc(3cd)i. 

The  radical  sign  means  that  the  root  of  the  quantity  under  it  is  to 
be  extracted ; which  may  be  done  by  extracting  the  root  of  the  several 
factors  successively:  whence  it  follows  that  we  may  extract  the  root 
of  any  factor  under  the  radical,  and  write  the  root  so  found  as  a 
factor  without;  thus, 

2aV  b'2c3md  = 2aV  b*<?md“d, 


2 «V  12hVW  = 2a x 3bVcd, 

= 4 abcy  3cd. 

This  operation  .is  called  simplifying  a radical.  A radical  is  said 
to  have  its  simplest  form  when  there  is  no  factor  under  the  radical 
of  as  great  a power  as  the  degree  of  the  radical.  Practically, 

To  simplify  a radical  expression,  look  out  all  the  factors  under  the 
radical  sign  vjlnch  are  exact  powers  of  the  degree  of  the  root  required, 
or  may  be  made  so  j divide  the  exponents  of  such  factors  by  the 
index,  and  write  the  factors,  with  their  new  exponents,  without  the 
radical. 

Or  if  the  parenthesis  is  used  instead  of  the  radical  sign, 


46  ELEMENTS  OF  ALGEBEA. 

Multiply  the  exponent  of  the  factor  by  the  exponent  of  the  paren 
thesis. 

Examples. 


1.  2 a(WcY~  =2 ax  2b(c)  2 = 4ab(c)i. 

2.  2aV Wc  — 2 a x 2 bVc  = 4a5 \/ c. 


5.  5x(xi(a+b)'j 3 — 5x3(a  + bf . 


6.  5xVxi{a  + b)  — 5 x^'s/a  + b. 

7.  3a  (d jAy {a + b)  - ~ = 9 az'(a  + b)y^. 


8.  7a V 1 6xy\a  + b )4  = 28a2y(a  + J)}v  x. 

9.  c(18«3&5)“  = c( 2 x 9 a2ab4b)~  — 3ab'2c(2ab)*~. 

10.  |V48 «m+2  = ?,V3  x 16«"‘fts  = 2«V3«”. 

11.  [32ft7(2  + J)4p=  [4  x 8«6fl(2  + b)\2  + b)^=2a\2  + b)  [4 a (2  + 6)] 

12.  \/ll2a3—4aV7a,  V ^°-.am+3=l-aV am+l, 


/324  _ 6 /3 

27a3“3«y  a 

13.  'V/f^«7 5 = ?pr \Tab,  = fr' Mar',  = \fa-\Taf‘. 


14. 


2 7ft  4 5 
Scd3 


b b 


_5 

b3‘ 


15.  vf=i\/3,  V&=Wb  W&=Wb 

16.  VW= Vw= Vfi  x io=t  ^10. 

17.  V-V  = ^11=  VoV  x 63  — y a/63. 

is.  vn=vj^M=wi 


MONOMIALS — EXPONENTS  AND  THE  SIGNS  X AND  -f-. 


47 


48.  To  pass  a Factor  under  the  Radical. 

It  is  evident  that  we  may  pass  a factor  from  without  to  within  a 
parenthesis,  or  the  radical  sign  by  the  reverse  process ; that  is,  by 
dividing  the  exponent  of  the  factor  by  the  exponent  of  the  paren- 
thesis, in  the  one  case ; and  by  multiplying  the  exponent  of  the 
factor  by  the  index  of  the  radical,  in  the  other. 


Examples. 

1.  3ff(6ff)^=(54«3)2. 

2.  2aV ab—2V a3b,  3c\koa4:d=  's/ 135 aic3d. 

3.  2ff(rt  + Z>)3  = 2(«3(ff  + £))3,  8a3(a— b)3  = (2a(a  — l?))3. 

4.  5a2\/cc  + b—5Va4:(a  + b),  5aVa—b=aV25(a—b). 

K + /27«3(rt  + £)\L 

b \d)  ~ \b2  ‘ d)  ’2  o\  c ) ~\  8c4  ) • 

6.  3«2(fl25mc-1)2  = (3%(«25’"c-,))2,  2arb"(j)™=  (— 

mm  m vi  p qm  p 

8.  cin(a)n  =(a2)T,  «n(«)?  = (ffps+I)7. 

9.  cu  k/ aT— 


p / pm 


~+T,  am+'mya=  V«(m+I)(m_1)+l. 


10.  bm-\b)^i=(b^+1)^ i,  ar(cr$L=(ari7'*-1)^. 


49.  To  Transform  the  sum  of  two  or  more  Radicals. 

Eadicals  are  said  to  be  Like  or  Similar  when  they  have  the  same 
indices  and  the  same  quantities  under  the  radical  signs ; thus, 
aVb  and  kVb  are  like;  (a  + b)^/amb  and  da2Va^b  are  like;  so  also 
. 3/ a— b , „ . / a—b 

°re  y — *ni  °j/ ■ 

The  algebraic  sum  of  two  or  more  like  radical  expressions  may 
readily  be  transformed  into  a single  term;  thus, 

3 a*\/ be a2b  *Jbc — 2\^bc={3a — a2b—2)'\/bc. 


48 


ELEMENTS  OE  ALGEBRA. 


We  have  simply  to  take  the  algebraic  sum  of  the  quantities 
without  the  radicals,  and  after  this  write  the  common  radical. 

If  the  quantities  without  are  numerals  entirely,  they  are  actually 
added  or  subtracted  according  to  their  signs  ; thus, 

3 Va^+l)  + 5 */a  + b = 8 v/«~+£- 

yiVH/i- 

Radicals  which  are  not  like  as  given,  may  sometimes  be  made  so ; 
thus, 

a^/^c—bVbcd2  = abVbc—bd\/bc 
— (ab—bd)Vbc. 

This  operation  of  reducing  radical  expressions  to  a single  term  is 
generally  called  addition  or  subtraction  of  radicals,  as  the  case  may  be. 

Examples. 

1.  3Va  + 2Va  = 5V~a,  aVa  + bVci=(a  + b)\/a. 

2.  8a2X'V/3c2  + 4a9z*/3c*  = 12a2x\/3c2,  oa^/b  — hb^/b  = 
{2>a—5b)Vb. 

3.  5a{b)^-2a(b)'2  = 3 a(b)^,  3(a  + S)2-  (a  + b)$=  2 (a +3)*. 


6.  2aVb^c—3bi\/bc^ =2ab'\/bc—3bc/\/bc=(2ab—3bc)^bcI 

7.  4(«3^3)'-  !(«&)*=  iab(ab)^-i((tb)i=(^ab-^){ab)k 

8.  9^l6^-10^54rt-1=18^^1-30^2^  = -12'V/2«i:i. 


MONOMIALS — EXPONENTS  AND  THE  SIGNS  X AND 


49 


10. 


a2b  la  nr  5a  n-  31a  rr 

T=8-.V»+3V?=1TVS: 


11.  Vi+VA=iV3  + iV3=fV3. 

13'  \/ 1 +i^=i^a- 


50.  General  Principles  of  Exponents. 

From  the  nature  of  exponents  we  have  the  following  general  re- 
sults : 


amXan—am+n 


— =am_n 
an 

(am)n=amn 

f/am=an. 


Thus,  we  may  say  that, 

I.  The  addition  of  exponents  takes  place  in  the  multiplication  of 
quantities. 

II.  The  subtraction  of  exponents  takes  place  in  the  division  of 
quantities. 

III.  The  multiplication  of  exponents  takes  place  in  the  formation 
of  powers. 

IV.  The  division  of  exponents  takes  place  in  the  extraction  of 
roots. 


50 


ELEMENTS  OF  ALGEBRA. 


SECTION  V. 

TRANSFORMATIONS  OF  POLYNOMIALS. 

51.  We  have  now  pretty  well  disposed  of  tlie  operations  upon  mo- 
nomials or  single  terms  ; let  us  proceed  to  investigate  the  transform- 
ations which  may  be  performed  upon  polynomials.  If  the  prin- 
ciples already  established  are  carefully  borne  in  mind,  we  shall  find 
little  difficulty. 

52.  Multiplication  of  a Polynomial  By  a Monomial. 

First,  let  a be  the  sum  of  all  the  positive  terms  in  any  polynomial, 
and  b be  the  sum  of  all  the  negative  terms  in  the  same ; the  polyno- 
mial itself  will  be  a— b. 

The  product  of  this  polynomial  by  any  single  term,  as  c,  will  be 

(a  — b)c. 

Let  us  now  convert  this  product  into  an  algebraic  sum.  The  prod- 
uct of  a and  c is  ac ; but  this  is  greater  than  the  true  value  of  the 
given  expression,  ( a—b)c , since  b should  have  been  taken  from  a be- 
fore it  was  multiplied  by  c.  If,  then,  we  subtract  be  from  ac,  we 
shall  have  the  true  product ; hence, 

(a—a)c—ac—bc. 

We  should  have  obtained  the  same  result  by  simply  multiplying 
each  term  of  the  polynomial  by  c,  observing  the  law  of  signs,  and 
uniting  the  partial  products  by  their  respective  signs. 

If  c had  been  negative  we  should  have  found  the  same  principle 
to  apply. 

We  may  say,  then,  in  general,  that, 

To  multiply  a polynomial  by  a monomial,  multiply  each  term  of 
the  polynomial  by  the  monomial,  remembering  that  like  signs  give 
plus,  and  unlike  minus. 


1.  5«2(3«5  + 2e—  1). 

2.  \a^{fa^b~x  + $). 

3*  3\/a(5\/ 6 — V«). 


Examples. 


Ans.  15«3Z>  + lOrt’c— Sab 
x 

Ans.  ab  +£a  • 
Ans.  15^/6# — 3a. 


TRANSFORMATIONS  OF  POLYNOMIALS.  51 


KI-v^h) 

. a°  a , a 

Ans ' i’—b^+Tv 

Vc(j/c-20  + 2\/c). 

Ans.  fyd— 20y/c  + 2c. 

—25a-xi(l  — 3abx~1  — a2). 

Ans.  — 2 Wa?x*  + rtha?bod  + 25  aJx*. 

2 a2[a  + b 3 «-s). 

Ans.  2a3  + 2a*b  6. 
c 

Va/Va  \/a\ 

1r~  a vV 

Vb\Vb  Vd 

Ax  /be,  • — £ 

h Vbd  J/b 

We  are  said  to  factor  an  expression  when  we  transform  it  so  that 
the  several  factors  which  enter  it  are  made  visible  to  the  eye ; thus, 
ab  + ac—ad—a{b+c—d).  When  we  have  an  algebraic  sum  in  which 
a monomial  and  a polynomial  factor  enters,  the  operation  of  factor- 
ing is  the  reverse  of  that  in  the  above  examples. 

Let  the  student  be  required  to  factor  the  answers  given  in  the 
above  examples. 


53.  The  Multiplication  of  Polynomials. 

Resuming  the  polynomial,  a—b,  let  c—d  be  any  other  polynomial. 
Their  product  will  be 

(a—b){c—d). 

To  transform  this  product  into  an  algebraic  sum,  let  us  begin  by 
multiplying  a— b by  c;  we  shall  have 

ac—bc. 

This  result  is  d times  greater  than  the  product  of  the  two  poly- 
nomials, since  d should  have  been  subtracted  from  c before  we  used  it 
as  a multiplier.  If  we,  then,  subtract  from  ac—bc  the  quantity  d{a  — b), 
we  shall  have  the  true  product  of  the  polynomials  ; that  is 

(a—b)(c—d)=ac—bc—(ad—bd)=ac—bc—ad  + bd. 

It  will  be  observed  that  we  have  simply  multiplied  each  term  of  one 
polynomial  by  every  term  of  the  other.  No  mention  was  made  of  the 
signs  in  deducing  this  result,  so  that  by  inspection  we  see  again,  that 
like  signs  give  plus,  and  unlike  minus. 

We  may,  then,  say,  that 


52 


ELEMENTS  OF  ALGEBRA. 


To  multiply  one  polynomial  by  another,  multiply  each  term  of  the 
one  by  every  term  of  the  other,  and  simplify  the  result. 

It  is  sometimes  more  convenient  to  set  one  polynomial  under  the 
other,  and  write  like  terms  under  each  other  as  the  operation  pro- 
ceeds; thus, 


a2  — ab  + b2 
a2  + ab  + b 2 

al—a3b  + a2b2 

a3b—a2b2  +ab3 

a2b2  —ab3  + bi 

«4  +a2b2  +b 4 

Examples. 


I.  (3a-2b2c-l)(ha2c  + b-of). 

Ans.  15a3  c— 10a2  b2c2  — oa2c+oab—2b3c—b—oax+2b2cx  + z. 


3.  (Va  + }Vb— i)(Va— \Vb  + \).  Ans.  a—\b  + \Vb— 


r.r,2  1 2.  2.  7 

— a2b— \ + c2  + a2bc3  c*. 

b- 

Ans.  JS+^l+^Z t«_l. 


c 


6.  (a  + b)(a— b). 

7.  («  + £)(« + 5)- 

8.  (a—b)(a—b). 


Ans.  a2—b2. 


Ans.  a2  + 2ab  + b2. 


Ans.  a2  —2ab  + b2. 


b 


Vbd  d' 


TEANSFOBMATIONS  OF  POLYNOMIALS. 


53 


i 


i 


(V«  + %/ab—  V~b)(^Va—  /\/~ab  + Vb')- 


Ans.  a-'^H2  +2^/a2b3  +b. 


14.  (— p + Vq+pz){  — p—  Vq+i>2)- 


Ans.  —q. 


54.  Two  Terms  unchanged  by  Simplifying’. 

A polynomial  is  said  to  be  arranged  with  respect  to  the  descend- 
ing powers  of  a particular  letter,  when  the  first  term  contains  that 
letter  with  its  highest  exponent,  and  the  next  contains  it  with  its 
next  highest  exponent,  and  so  on.  It  is  arranged  according  to  the 
ascending  powers  of  the  letter  Avlien  the  order  of  terms  is  reversed. 

Now,  when  two  polynomials  are  multiplied  together,  there  are 
always  two  terms  in  the  result  which  cannot  disappear  in  the  pro- 
cess of  simplification.  These  are,  first,  the  term  which  results  from 
combining  the  two  terms  with  the  highest  exponents  of  a particular 
letter ; and,  second,  the  term  which  results  from  combining  the  two 
terms  with  lowest  exponents  of  the  same  letter;  thus, 


In  this  case,  x5  and  — 2x2  come  immediately  from  the  combina- 
tion of  certain  terms;  while  the  others  are  the  results  of  simplifying. 

55.  The  Division  of  Polynomials. 

The  dividend  must  always  be  produced  by  multiplying  the  quotient 
by  the  divisor ; and  when  these  are  polynomials,  we  know,  from  the 
last  article,  that  there  must  always  be  two  terms  at  least  in  the  divi- 
dend which  will  undergo  no  change  from  the  process  of  simplifica- 
tion. 

Let  us  take  the  polynomial  x5  — 5.r4  + 7x3  — 2x2,  and  let  the  divisor 
x2—2x  be  given  to  find  the  quotient. 

Now  since  x5  must  have  resulted  from  the  multiplication  of  x2  in 
the  divisor  by  that  term  in  the  quotient  which  contains  the  highest 
power  of  x,  we  can  tell  at  once,  by  dividing  x5  by  x2,  what  that  term 
must  have  been.  One  term  of  the  quotient  must  then  be  x3. 


(x3—3x2  +x)(x2—2x)  — 
x 3 — 5x 4 +7x3  —2x2. 


51 


ELEMENTS  OE  ALGEBRA. 


Let  us  for  convenience  arrange  both  polynomials,  and  write  the 
dividend  first  and  the  divisor  after  it ; thus, 

x5  —ox4  + 7a;3  —2a;2  x2  —2x 
x 5 — 2a;4  a;3 — 3x2+x 

—3x4  + Hx3—2x2 
— 3a;4  + 6a;3 

x3—  2x2 
x3  — 2a:2 

0 

writing  the  first  term  of  the  quotient  under  the  divisor.  Now,  as 
this  term  of  the  quotient  must  be  multiplied  into  every  term  of  the 
divisor,  in  the  process  of  forming  the  dividend  by  multiplying  the 
divisor  and  quotient  together,  if  we  perform  this  multiplication  and 
subtract  the  result  from  the  dividend,  we  shall  free  it  from  all  partial 
products  in  which  this  term,  x2,  of  the  quotient  enters.  Multiplying 
and  subtracting,  as  shown  above,  we  have  a remainder,  — 3a;4  + 7a;3 
-2a;2. 

Now,  the  term  —3a;4  of  the  remainder  must  have  resulted  from 
multiplying  x2  by  that  term  of  the  quotient  which  has  the  next 
highest  power  of  x in  it.  We  may  thus  find  another  term  of  the 
quotient  by  dividing  the  first  term  of  the  remainder  by  the  first  term 
of  the  divisor;  thus  we  have,  — 3.r2.  Writing  this  after  the  term  of 
the  quotient  already  found,  and  multiplying  the  divisor  by  it,  as  be- 
fore, and  subtracting  the  result  from  the  remainder,  we  have  a second 
remainder,  x3  — 2x2.  By  continuing  this  process,  we  shall  find  all 
the  terms  of  the  quotient.  If  there  should  prove  to  be  a final  re- 
mainder, the  division  cannot  be  exactly  performed.  In  this  case,  we 
may  write  the  remainder  in  the  form  of  a fraction,  having  the  divisor 
for  the  denominator,  and  unite  it  with  the  quotient  by  its  proper 
sign. 

We  may  then  say,  practically,  that 

To  divide  one  polynomial  by  another',  arrange  the  dividend  and 
divisor  according  to  the  powers  of  the  same  letter ; divide  the  first 
term  of  the  dividend  by  the  first  term  of  the  divisor.  This  will  be  the 
first  term  of  the  quotient. 

Multiply  the  divisor  by  th  is  term  of  the  quotient  and  subtract  the 
result  from  the  dividend. 

Divide  the  first  term  of  the  remainder  ( arranged ) by  the  first  term 
of  the  divisor  ; multiply  as  before , and  subtract  from  the  remainder. 


TRANSFORMATIONS  OF  POLYNOMIALS. 


55 


Continue  the  operation  until  there  is  no  remainder,  or  until  the  first 
term  of  the  remainder  will  not  contain  the  first  term  of  the  divisor. 


Examples. 

1.  (a2A2abAb2)-A(aAb). 

2.  (a2—  2abAb2)A-(a— l). 

3.  (a2—  b2)A-{aAb). 

4.  (4«3  4-4«2— 29«4-21)-4-(2a— 3). 

5.  {a^-b^Afi-b). 

6.  (xs  Ays)A-(xAy). 

7.  (x3  Ay3)A(x2  + y2). 
a3-b5 


8. 


A ns.  aAb. 
Ans.  a—b. 
Ans.  a—b. 
Ans.  2a2  A 5a  — 7. 
Ans.  a3  +a2b  + ab2  +b3. 
Ans.  xi—x3yAx2y2  — xy3Ayi. 

Ans.  xi—x2y2+yi. 

Ans.  a4  +a3b  + a2b2  Aab3  + bi. 


am—bm 
a — b 

10  12^~192 
• 3 y-5  ' 

11.  l-4-(l-«). 


Ans.  am~' + am~sb  + am-3b'3  + ■ • • bm 


Ans.  4?/ 3 4- 8?/ 2 4-16^/4-32. 


Ans.  \ A a A a2  A a3  A a^  A 


12. 


-fia2  +4  V a3  +8'V/a  + 8v/« 


\/a  + 2 'fia 


13. 


14. 


15. 


a + a2  b2  + b 

~x  i i I ' 
a2  4- ft4#4  + b 2 


fa  e\  f's/a  \fic\ 
\h  d)  ' ~ rfj)' 


A A 'AA  tA  A 1C 

a 4 + b^ 


Va  — fib 

' 'fia- 'fib 


1 —a 

Ajis.  Va  + 2^+4. 


1 Li  i 

Ans.  a2—a‘Lbi  Ab2. 


'fia  'fic 

Ans.  —n  4 — , 

'fib  fi  d 


L i 

Ans.  «4  A 'A. 


Ans.  fia  A ‘fib- 


56 


ELEMENTS  OF  ALGEBRA. 


56.  Formulas  used  in  Transformations. 

A Formula  is  any  general  truth  expressed  by  means  of  symbols. 

The  transformation  of  polynomials  can  often  be  much  shortened 
by  the  use  of  certain  simple  formulas.  We  shall  now  give  a few  of 
these. 

(a  + b)2  — {a  + b){a  + b)=a2  + 2 ab  + b2  ; that  is. 

The  square  of  the  sum  of  tivo  quantities  is  equal  to  the  square  of 
the  first,  joins  twice  the  product  of  the  first  and  second,  plus  the  square 
of  the  second  ; and  the  converse. 

(a—b)2  = (a—b)(a—b)  — a2—2cib  + b2  ; that  is, 

The  square  of  the  difference  of  tivo  quantities  is  equal  to  the  square 
of  the  first,  minus  twice  the  product  of  the  first  and  second,  joins  the 
square  of  the  second  ; and  the  converse. 

(a  + b)(a  — b)=a2  — b2, 

The  product  of  the  sum  and  difference  of  tivo  quantities  is  equal  to 
the  difference  of  their  squares,  and  the  converse. 

When  the  second  member  of  an  equation  is  the  algebraic  sum  re- 
sulting from  operations  indicated  in  the  first,  such  sum  is  called  the 
development  of  the  first  member;  thus,  in 

(a  + b)2=a2  +2  ab  + b2, 

the  second  member  is  the  development  of  the  first. 


Examples. 


Develop  the  following  expressions, 
formulas. 

by  the  aid  of  the  foregoim 

1.  {x  + y)2. 

Ans.  x2 +2 xy+y2. 

2.  (2a  + 2b)2. 

Ans.  4r<2  +8u5  + 452. 

3.  (a  — l)2. 

Ans.  a2—  2a + 1. 

4.  (i ~a)2. 

Ans.  \—a  + a2. 

5.  (\a+\b)2. 

Ans.  }a+^ab+£b2. 

„ fVa  ,\2 

. a 2 \/a  , 

6.  [ A—  — 1 ). 

Ans.  y 1^+1. 

' V b o 

t Vb 

TRANSFORMATIONS  OF  POLYNOMIALS. 


57 


7. 

\(a-b)s  J 

8.  (p—  Vq+p2)2- 

9.  (1  -a~iy. 

10.  (x  + y)(x-y). 

11.  (£a— b)(^a  + b). 

12.  (3a-i-b)(3a  + ib). 

_ (a  c\(a  c\ 

13-(i  + S 

14.  (V«  + 5)(a/«  — b). 

15.  (1— a)(l + a). 

16.  (l-i«)(l  +*«). 

17.  («2“  + 54m)  (a2"1— bim). 

18.  (‘\fa—  v'i)(v/a  + y7>). 


^4  as. 


a + Z> 
a— Z> 


2a^(a  + Z>)^ 
(a— b)2 


+a. 


Ans.  2)‘z  — 2p‘\/q+p)2  + {q+p2) 
Ans.  1—2  a_I  + a~2 
-d?2s.  a:2 — y 2 


JhiS.  Jft2— 52 
^4?is.  9a2— JJ2 


-4^5.  TIT 7T 


62  d2 


Ans.  a—b 
Ans.  1— a2 
Ans.  1— ia2 
Ans.  aim—b8m 


Ans.  'i/a2  — 


57.  Formulas  used  in  Transformations—  Continued. 

By  developing  the  following,  we  have 

(i a + b)(a  + c)—a 2 + (b  + c)a  + bc 
(a—b){ci—c)  — a^  — (b  + c)a  + bc. 

"Whence  we  see  that  when  two  binomials  are  to  be  multiplied  to- 
gether, having  their  first  terms  the  same  and  the  second  terms  differ- 
ent, with  the  same  middle  signs,  we  may  write  out  the  result  at  once: 

1.  The  first  term  of  the  result  must  be  the  square  of  the  common 
term ; 

2.  The  second  term  must  be  the  numerical  sum  of  the  two  last 
terms  into  the  first  term  with  the  sign  of  the  last  terms;  and 

3.  The  last  term  must  be  the  product  of  the  last  terms  with  the 
plus  sign. 


Examples. 

1.  (a  + 2)(a  + 3)=a2  + 5a  + 6. 

2.  (x— 5)(z— 4)=.-r2  — 9£  + 20. 

3.  (1  +a)  (1+ b)— 1 + (a  -|-  V)  + cib. 


58 


ELEMENTS  OF  ALGEBRA. 


4.  (a25  + 5)(«26  + 7)  = «452  + 12a2&  + 35. 

5.  (3a-J-)(3a-i)=9a2-fffl  + ^. 


/Va 

wf 


'V  a 

Wl 


a2 

11  a 

~ ¥ 

+ ~b 

—20j 

i-f  : 

b 

18. 

21  '/a 

Vb 


+ 20. 


8.  (a-2— l)(a-2— o)=a~Ji— 6«_2  + 5. 


58.  Formulas  used  in  Transformations — Continued . 

Now  let  the  last  terms  be  different  with  different  signs ; thus, 

{a  + b){a  — c)  = a2  + (b—c)a—bc. 

Whence  we  see  that  in  this  case  the  middle  term  will  be  the  differ- 
ence of  the  last  terms  into  the  common  term,  and  the  last  term  will 
have  the  — sign. 


Examples. 

1.  («  + 3)(ffl— 5)=a2—  2a— 15. 

2.  (a*-9)(a*+3)=a-6o*-27. 

3.  (3a2Z>4  — 2)(3a254  + 8)  = 9a4£8  — 18«254  — 16. 

4.  (\/a— l)(V«+i)=«— \Va— h 

5.  (/\/~x—5)(\/ x + V)=  \/x2 +2\/ x—3o. 

6.  (a_1  + 9)(a_1  — 11)=«-2— 2a~l  — 99. 

7.  (am  - 4)  (am  + 12) = dim  + 8 am — 48. 

m m 2to  to 

8.  (xn  — 9)(^n  + 2)=a:+  — 7%n  — 18. 

9.  (ai-l)(«f  + 25)=a^  + 24al-25. 


59.  Factoring:. 

By  the  aid  of  the  simple  formulas  already  explained,  we  may  often 
resolve  trinomials  into  their  factors  by  mere  inspection.  Let  us  first 
take  some  examples  under  the  three  following  formulas: 

a~  +2  ab  + b2  — (a  + b)(a  + b). 


TRANSFORMATIONS  OF  POLYNOMIALS. 


59 


a2—2ab  + b2  — (a—b)  (a—b). 
a2—  b2  — (a  + b)(a—b). 

Examples. 

1.  a2b2 —c2  = (ab  + c){ab—c). 

2.  4a2—  4a  + l = (2a— 1)(2«— 1). 

3.  z2  +z+l=(z+%)(z+l). 

5.  a— 5=(V«+  Vb)(Va—  Vb. 

6.  4«2— i=(2a— + 

7.  25 a2bi  +20ab2c  + 4c2  — (5ab2  + 2c){5ab  + 2c). 

0 25a-2  x_/'oa~ 1 f oer1 

^b^~y-{jr+^j/^b 

9.  J4  = (ai+ 

10.  V .r— j=(v/.t+J)(  v^— 


60.  Factoring—  Continued. 

Let  us  now  show  the  operation  of  factoring  under  the  formulas, 

(i a + b){a  + c)—a 2 + ( b + c)a  + bc , 

(a—b)(a—c)=a2  — (b+c)a  + bc. 

For  example,  take 

a2  - f-  5a  -f-  6. 

If  there  are  any  binomial  factors  in  this  expression,  the  product  of  their  first 
terms  must  produce  a'2,  and,  since  a enters  the  middle  term,  they  must  both  be 
a : since  the  last  term  is  positive  the  last  terms  must  have  like  signs.  Looking 
at  the  middle  term  we  see  that  they  must  be  + ; we  may  then  write  so  much  of 
the  required  factors  ; thus, 

a+  , a+  . 

Now  we  must  have  two  such  quantities  for  the  second  terms,  that  when  mul- 
tiplied together  they  will  produce  6,  the  last  term  of  the  trinomial,  and  when 
added  will  give  5.  These  terms,  then,  can  only  be  +3  and  +3,  and  hence  the 
factors  are  a + 2 and  a + 3,  and  we  have 

a2  4-  5a  + 6 = (a  + 2)(a  + 3). 


60 


ELEMENTS  OF  ALGEBRA. 


Again,  take 


a2— 5a + 6. 


Here  the  middle  term  being  — , the  last  terms  of  the  factors  must  be  — also  ; 
thus. 


a2  — 5a+6=(a— 2)(a— 3). 


When  the  last  sign  in  the  trinomial  is  — , thus  ; 

a2+2a— 8, 


the  last  terms  must  be  unlike  to  give  a — quantity,  and  the  + must  be  the 
greater  to  give  +2  in  the  middle  term.  Those  terms  must  thus  be,  +4  and 
—2,  and  we  have, 

a2+2a— 8=(a+4)(a— 2). 

Examples. 

1.  x2  + 3x  + 2=(x+2)(x-\-l). 

2.  x2  + 11x  + 72=(cc  + S)(x  + 9). 

3.  x2  — 12x  + ?>o  — (x— 7)(x— 5). 

4.  x2  +2ix—2o—{x  + 2b){x—l). 

5.  x2—  f£  + i=(.r— 4)(a— J). 

6.  a2x2  — 74ax—  75  = (a.T  + l)(«.r— 75). 

7.  9rt2+12a— 5 = (3«  — 1)(3«  + 5). 


10.  \/a— 3-v/«  + 2 = (\/a— 1)(  v'ff— 2). 

11.  «_2m  + 5a~m  + 6 = (ar®  + 3)(a~m  + 2). 

12.  25x2  — 60j:  + 35  = (5x2  —7)(5x2  — 5). 

Remark. — When  there  is  a monomial  factor  present,  remove  it  first  and  then 
factor. 


13.  a2x2-l2a2x  + Zba2=a2{x-7){x-o). 

14.  5«2+15a  + 10  = 5(rt  + l)(«  + 2). 

15.  2x2y2-^^-  + ^|~  = 2t/2(.r-i)(.T-i). 


16.  a2 \To— ZaVb  + 2^/b=  Vb(a  — l)(fi  — 2). 


TRANSFORMATIONS  OF  POLYNOMIALS. 


61 


61.  The  Division  of  the  Difference  of  like  Powers. 

The  difference  of  the  like  powers  of  any  he o quantities  is  always 
divisible  by  the  difference  of  the  quantities  themselves. 

For,  let  a and  b be  any  two  quantities,  and  m be  any  positive  whole 
number.  The  difference  of  the  like  powers  will  be  am—bm. 

Beginning  the  division,  we  have, 

am—bm  a—b 
am—am~'b  a"1-1 

am~xb—bm,  or 
b{am~x-b^). 

Now,  if  this  remainder  is  exactly  divisible  by  a—b,  then,  oT—bm  is 
itself  exactly  divisible  by  it.  But  the  remainder  is  b times  am~l— 
bm~\  so  that  if  the  factor  am~x—bm-'  is  exactly  divisible  by  a—b,  the 
entire  remainder  is  divisible  by  this  difference,  and  hence  the  divi- 
dend am—bm  is  likewise  so  divisible.  We  see,  thus,  that  if  am~x—bm~x 
is  divisible  by  a— b,  am— bm  is  also  divisible  by  the  same  quantity; 
that  is  to  say,. 

If  the  difference  of  the  like  powers  of  two  quantities  is  exactly  di- 
visible by  the  difference  of  the  quantities  themselves,  then,  the  differ- 
ence of  such  powers  greater  by  unity,  is  also  exactly  divisible  by  the 
difference  of  the  quantities. 

But  we  know  that  a~—b~  is  exactly  divisible  by  a—b;  hence, 
from  the  hypothetical  proposition  just  established,  a3—b 3 must  like- 
wise be  so  divisible,  and  hence  «4 — d4  must  be,  and  so  on  to  in- 
finity ; which  was  to  be  proved. 

This  method  of  proof  is  called  Mathematical  Induction. 

The  form  of  the  quotient  will  be, 

ft711 Iff1 

— —ar-^Jram-ib  + am-3bz+ + 

a—b 


62.  The  Division  of  like  Powers. 

The  following  propositions  may  also  be  readily  demonstrated: 

1.  The  differejice  of  the  like  even  powers  oftivo  quantities,  is  always 
divisible  by  the  sum  of  the  quantities.  The  form  of  the  quotient 
will  be, 

/,2m l-2m 

- 'b  + a?m~*b  + • • • • -J*— *. 

a + b 

/y4 7i4 

f - =«3  -a*b  + ab* -b3. 

a + b 


62 


ELEMENTS  OP  ALGEBRA. 


2.  The  sum  of  the  like  odd  powers  of  two  quantities  is  always  di- 
visible by  the  sum  of  the  quantities  themselves.  The  form  of  the 
quotient  will  be, 

=a!m — a2"*_1£  + a2m_2Z>2—  • • • + J2™. 

a + b 

a \ =ai  — a3b  + a2b2  — ab3  + bi. 
a + b 


Examples. 

Write  out  the  developments  of  the  quotients  indicated. 

x2—y2  x3  — y3 

1.  — —x  + y,  — =x2  +xy+ y2. 

x—y  J x~y 


n x3  + y3  „ „ xi  — yi 

2.  : — =x2—  xy  + y2,  — 

x + y J J x + y 

Q/y37.3 ^3 

3.  — = 4a2£2  + 2abc  + c2. 

2ab—c 

4.  -~+  '’2  = a4  -2a3  + 4a2  - 8a  + 16. 

a + 2 

5.  Factor  aG—bc,  x3  — l,  1— a’3,  8a3— 27. 

6.  a5b—9b3,  12x4-192, 


-x3  —x2y  + xy 2 —y3. 


7.  a~3-b-3, 


8. 


ai  — bi  (a2  —b2)(a2  +b2)  (a— b)(a  + b)(a2  +b2)' 

a8—b3  _(a4  — bi)(ai  +bi)  (a2  + b2)(a2  —b2)(a4  + bi)  _ 


''  a2 + 2ab  + b2  ( a + U)(a  + b ) 

( a 2 + b2)(a  + b)(a—b(a4  + 1 >i) 

(a  + b)  (a  + b ) 


(a  + b)(a  + b) 


63.  The  Binomial  Formula. 

The  process  of  raising  a binomial  to  any  power  may  be  greatly 
shortened  by  using  what  is  called  the  Binomial  Formula.  By  actual 
multiplication  we  have  the  following  developments: 

1st  Power  (a  + b)  =a  + b 

2d  “ (a  + b)2=a2 +2ab  + b2 

3d  “ (a  + b)3=a3 +Za2b  + 9ab2  +b3 


63 


TRANSFORMATIONS  OF  POLYNOMIALS. 

4th.  Power  (a  + Z»)4=«4 +4a35 + 6a253 +4«53 

5th  “ (a  + b)5  —a5  + 5aib  + 10a3b2  + 10a2Z»3  +5«54  + b5. 

Examining  any  one  of  these  developments,  as  the  last,  we  see  that 
a (called  the  leading  letter)  appears  in  the  first  term  with  an  expo- 
nent equal  to  the  exponent  of  the  power  of  the  binomial  in  that  case, 
and  that  this  exponent  goes  on  diminishing  by  unity  to  the  last  term, 
in  which  it  enters  to  the  zero  power,  that  is,  not  at  all;  thus, 

a5  o4  a3  a2  a1  a°. 

With  b the  order  is  just  reversed ; thus, 

b°  b1  b2  b 3 54  b5. 

It  will  be  observed,  further,  that  any  numerical  co-efficient  may  be 
found  from  the  preceding  term  by  multiplying  the  co-efficient  of 
that  term  by  the  exponent  of  a,  the  leading  letter  in  that  term,  and 
dividing  this  product  by  the  number  of  terms  preceding  the  re- 
quired term ; thus,  to  find  the  co-efficient  of  the  fourth  term,  in  the 
development  of  the  fifth  power  above,  multiply  10,  the  co-efficient  of 
the  third  term,  by  3,  the  exponent  of  a in  that  term,  and  divide  the 
product  30  by  3,  the  number  of  terms  preceding  the  fourth,  and  we 
have  10,  the  co-efficient  required.  In  like  manner,  the  co-efficient 
of  any  other  term  may  be  found,  remembering  that  the  co-efficient 
of  the  first  term  is  always  unity.  Generalizing,  we  may  write, 

(a+b)m —am  + mam~lb  + m ^ - ^-a'n~-b2  + 

— — pjTg ,-am~3b3  + +bm. 

This  is  The  Binomial  Formula,  and  it  may  be  rigidly  demon- 
strated to  be  true,  whether  m be  entire  or  fractional,  positive  or 
negative.  When  m is  fractional  or  negative,  the  number  of  terms 
will  be  infinite. 

The  power  of  any  binomial  may  be  developed  by  means  of  this  for- 
mula. For  example,  let  us  take, 

(3rc2  — 2«52)4. 

Let  ox2— a and  —2 ab2—b. 

We  have  at  once, 

(a  + b)i=ai  +4«35  + 6rf2Z>2  +4  ab3  + bl. 


64 


ELEMENTS  OF  ALGEBRA. 


Now  in  this,  writing  for  a and  b their  values  as  above,  we  have, 
(3a;2  — 2a&2)4  = (3a:2)4  + 4(3a;2)3(— 2a52)  + 6(3a;2)2(— 2a£2)2  + i(3z2) 
(~2a&2)3  + (-2aS2)4. 

(3x2—2ab2)i=81x8—216xeab2+216x±a2bi—96z2a3bs  + 16aibs. 

Examples. 

1.  Develop  (a2 +1)3,  («62-2)4,  (|+|)3. 

2.  Develop (V-1^  , (y/a-l)3,  (Va  + V^)4. 

3.  Develop  ( a + b )8,  (a  + ^^a^  + ^a^-1#— ^a^~2b2  +etc. 

4.  Develop  (a  + &)_1,  (a  + &)-1=a-1  — a~-b+a~3b2  — etc. 

5.  Develop  (a2'— l)3,  (V^+^J  , (x—y)x. 

6.  Develop  (2-3a;)-2,  (V^-V^)3. 

7.  Develop  (1— a)^,  (a  + 1)  3,  (a  + b)”. 


64.  The  Powers  of  Polynomials. 

The  power  of  any  polynomial  may  he  developed  by  the  use  of  the 
binomial  formula.  For  example,  let  it  be  required  to  find  the  third 
power  of  2a2  — 4aZ>4-3c2. 

Let  2a2  — ^ab—a  and  3c2  — b ; then 

(a  + b)3=a3  + 3a2b  + 3ab2  + b's. 

Replacing  the  values  of  a and  b,  we  have 
(2a2  - 4ni  + 3c2 ) ■ 3 = (2a2  - 4 ab) 3 + 3 (2a2  - 4 ab) 2 (3c2 ) + 3(2a 2 - 4a£) 
(3c2)2  + (3c2)3. 


Examples. 

1.  Develop  (a  + 5 + c)3. 

2.  Develop  (1— 2a;— 3»2)2. 

3.  Develop  (a’"  + 5’>)3. 

4.  Develop  (a2  + l+a~2)2. 

5.  Develop  (VTi+l  — V^)3- 


TRANSFORMATIONS  OF  POLYNOMIALS. 


65 


65.  Common  Multiple. 

The  Common  Multiple,  or  a Common  Dividend  of  two  or  more 
quantities,  is  any  quantity  which  is  exactly  divisible  by  each  of 
them:  such  a dividend  may  always  be  found  by  multiplying  the  sev- 
eral quantities  together  ; thus,  4«2,  'dab  and  3b2c  would  give  3 6a3bzc 
for  a Common  Dividend. 

The  Least  Common  Multiple  is  the  least  quantity  which  is  so  di- 
visible; thus,  V2a2b2c,  is  the  least  Common  Multiple,  or  Dividend, 
of  the  quantities  above  given. 

A common  multiple  must  obviously  contain  every  factor  which 
enters  any  of  the  quantities  by  which  it  is  to  be  divisible,  as  many 
times  as  such  factor  enters  any  one  of  the  quantities ; hence,  if  all 
the  factors  are  made  to  appear,  we  have  but  to  take  each  factor  the 
greatest  number  of  times  it  enters  any  one  of  the  quantities,  and 
multiply  the  results  together ; thus,  a2  + 2ab  + b2,  a2 —b2  and  a2  — 
2 ab  + b2,  resolved  into  their  simplest  factors  give, 

(a  + b)  {a  + b),  (a  + b)  {a—b),  {a  — b)  {a—b). 

Here  ( a + b ) enters  the  first  expression  twice,  and  ( a—b ) enters  the 
third  twice : the  second  expression  containing  no  other  factors  than 
these,  we  have, 

(a  + b)2(a-b)2=2a2  +2b2. 

It  is  commonly  better,  however,  to  retain  the  indicated  product  as 
in  the  first  member  of  this  equation,  than  to  develop  it  as  in  the 
second  member.  Thus,  wre  may  say  that, 

To  find  the  Least  Common  Multiple  of  two  or  more  quantities,  re- 
solve the  quantities  into  their  simple  factors,  and  take  each  factor  the 
greatest  number  of  times  it  is  found  in  any  one  of  the  expressions. 
The  product  of  these  factors  will  be  the  multiple  required. 


Examples. 

Find  the  Least  Common  Multiple  of  the  following : 

1.  a2—b2,  ab  + b2,  a2—ab,  { a + b )2,  {a  + b){a— b),  b{ci  + b),  a{a—b), 

{a  + b)2.  Ans.  ab{a-'-b)2  {a—b). 

2.  x2 +x—2,  x2 +2x— 3.  Ans.  (x—l)(x+2)(x  + 3). 

3.  {x2  — a)2,  x2—a,  5.  Ans.  5{x2—a)2. 

4.  x~  +Qx+20,  x2  + ox — 4,  x~  T 4a:  5. 

Ans.  (.r  + 4)(a:  + 5)(.T— 1). 


6G 


ELEMENTS  OF  ALGEBEA. 


5.  6a2— 23a + 7,  Ga2  —25a  + 14.  Ans.  (. 2a-7)(3a-l)(3a-2 ). 
G.  3a,  (a  + £)2,  a2  —2ab  + b2.  Ans.  3a(a  + b)2(a—b). 


66.  Operations  upon  Algebraic  Fractions. 

The  student  is  supposed  to  be  already  familiar  -with  the  manage- 
ment of  fractions  in  arithmetic.  Their  treatment  in  algebra  is  al- 
together the  same.  The  transformation,  however,  of  the  sum  or 
difference  of  two  fractions  into  a single  expression,  requires  some 
care,  and  we  shall  now  give  a number  of  examples  for  practice. 

Let  the  student  remember  to  first  convert  the  fractions  into  equiv- 
alent fractions  with  a common  fractional  unit,  and  then  to  add  or 
subtract  the  numerators,  placing  the  result  over  the  common  denom- 
inator. 


Examples. 

^ 1 1 _ a — l)  a + 1)  _a—b+a  + b _ 2a 

a + b a — b a 2 — b2  a2  — b2  a 2 — b2  a2 — b2  ' 

„ 1 2 _ x + 2 2(.r  + l)  _x  + 2— (2(a;  + l)) 

x + 1 x+2  (a;  + l)(rc  + 2)  (.r  + l)(a;  + 2)  (x  + l)(a;  + 2) 

—x 

(x  + l)(x  + :>)  ' 


1—  a 1 + a _(1— a){a  + b)  — (l  + a)(rc— b) 
+rZ-J+~la  + b)2~  (a-b){a  + b)~ 

a + b— a2  —ab  — a + b— a2  +ab  _ 2b— 2a2 

(a  — b){a  + b)2  (a—b)(a  + b)2 ' 


. 1 — a 1 + a 1 

4. 1 1 : 

1 + a 1 —a  (1  + a)2 

1 — 5 a — 4a2 


(1—  a2)(l  + a) ' 

_ a 2b  b 

5.  — -i j • 

a+b  a—b  a+b 


„ a + 3 2a— 5 

6.  — 1 — - 

o 3a 


1— a)(l— a2)  — (l+a)(l  + a)2  + 1— a 
( 1 + a)  (1  + a)  (1 — a) 


A ns. 


a + b 


A ns. 


a—b 

3a2  + 19a— 25 
15a 


a+b  a—b 
a—b  a + b' 


_ I ns. 


4 ab 

a2  — b2 


TRANSFORMATIONS  OF  POLYNOMIALS. 


67 


x I x—a\ 

V \X-—} 


8.  3x  + y 


9.  * + am~l 


10. 


am  + bm  ' aim—¥m 

1 1_ 

a*— I)*  a+  + bs 


u Va Vb 

\/a—Vb  y/a  + Vb 

12-  —h+lfr- 

a— b a+b 


Ans.  2 x+ 


Ans. 


cx  + bx—ab 
be 

2am—2bm 


aim_bim 


Ans. 


Ans. 


2b~ 
a—b ' 

a + b 
a—b ' 


Ans. r. 

ab 


67.  Essential  Sign. 

The  sign  by  which  one  expression  is  connected  with  another  is 
called  the  Sign  of  Operation  ; the  sign  resulting  from  the  combina- 
tion of  this  sign  with  the  sign  of  the  quantity  itself  is  called  the 
Essential  Sign  of  the  expression:  thus,  in  the  expression 

a—(—b), 

the  second  term  is  to  be  subtracted  from  the  first,  and  the  sign  of 
operation  is  — ; but  removing  the  parenthesis  the  two  — signs  com- 
bine and  give  +b.  This  resulting  sign  is  the  essential  sign  of  the 
term. 

We  sometimes  cannot  tell  the  essential  sign  of  an  expression  abso- 
lutely ; thus,  in 

a 

X — “7  , 

b—c 

the  essential  sign  of  the  fraction,  and  so  the  sign  of  x,  will  depend 
upon  the  relative  values  of  b and  c.  If  b>c  it  is  + ; if  5<c  it  is  — . 


Examples. 

1.  What  is  the  essential  sign  of  x in  x — 
tive  and  b>c?  when  5<c? 


a 


b-c 

a—b 


, when  a is  nega- 


2.  What  is  the  essential  sign  of  x in  x—  " — - , when  a>b  and 

c — ct 

c<d?  when  a<b  and  c<cl?  when  a<b  and  c>d?  when  a>b  and 
c>d? 


68 


ELEMENTS  OF  ALGEBKA. 


68.  Imaginary* Quantities. 

We  have  seen  that  the  even  root  of  a negative  quantity  is  imagi- 
nary. Let  us  now  take  an  imaginary  quantity  of  the  second  degree, 
as  V — and  multiply  it  by  itself,  applying  the  general  law  of  signs ; 
we  shall  have, 

V —a  x a/— a—  */af=  ±a. 

But  the  square  of  the  square  root  of  a quantity  is,  from  the  defi- 
nition, the  quantity  itself : so  that, 

{V— a)2  = — a. 

We  should  thus  have  —a=±a,  or,  taking  the  upper  sign,  — a— 
+ a,  which  is  impossible. 

To  avoid  this  difficulty  we  must  modify  the  law  of  signs  in  the 
multiplication  of  imaginary  expressions. 

Now,  every  imaginary  expression  of  the  second  degree  may  be  put 
under  the  form  of, 

a's/  — 1 : 

in  which  a represents  any  real  quantity,  whether  its  exact  value  may 
be  found  or  not;  and  a/— 1,  an  imaginary  quantity,  which  is  called 
the  imaginary  factor. 

For  example,  let  us  have, 

V — ab. 

This  may  be  written, 

V —ab=Vcib  x ( — 1 ) — ^/ab  . a/— T. 

Again. 

V— (a  + b) 2 = V(a  + b) 2.  V"— 1 = (a  + b)  a/— 1. 

Let  us  now  multiply  V — 1 by  itself  any  number  of  times,  until 
we  discover  the  law  of  such  combinations. 

Since  the  square  of  the  square  root  is  the  quantity  itself, 
a/— 1 x V — 1 = (a/ — l)s  = — 1. 

The  third  power  will  be  found  by  multiplying  the  second  by  the 
first;  hence, 

(a /^)2  x = = 

In  like  manner, 

( a/11!)  4 = ( 2 ( 2 = ( - 1)  ( - 1)  = + 1. 


TRANSFORMATIONS  OF  POLYNOMIALS. 


69 


We  may  in  the  same  way  continue  the  operation,  and,  for  the  sev- 


(y=I)3=(V^2(y=I)==--iy=I=  -V=i. 

(v^)4=(y^)2(v~i)2=(-i)(-i)=  +1. 

( v ^ ) 5 = ( v ) 2 ( v ^1 ) 3 - ( - 1 ) ( - v “y ) - v ^ • 

(y=I)6=(V^i)2(V^i)4-(-i)(  + i)=  -i. 

(V:zI)7=(V=i)6(a/:ii)=(-i)(V^i)=  -V^l. 
(v^iy=(v~i)i(v^;i)i={+i){+i)=  +i. 


It  will  be  observed  that  the  last  four  results  are  but  a repetition 
of  the  preceding  four,  and  so  they  would  continue  to  repeat  them- 
selves in  sets  of  fours. 

The  multiplication  of  imaginary  quantities  is  effected  by  the  use 
of  this  imaginary  factor ; that  is,  we  first  resolve  each  expression 
into  two  factors,  one  real  and  the  other  the  imaginary  factor,  V — 1 ; 
we  then  combine  the  real  factors  by  the  ordinary  laws,  and  the 
imaginary  factors  according  to  the  laws  just  deduced  for  the  forma- 
tion of  its  several  powers. 

If,  however,  only  one  of  the  quantities  is  imaginary,  the  ordinary 
rules  apply. 


eral  powers  of  V—l,  we  shall  have, 


(V— l)1  = 

(V^1)2  = 


V=L 


-1. 


Examples. 


1.  Multiply  5 a/— 4 by  2a /— 4. 


oa /iV” — 1 x 2 \/4a/  — 1 
10a/=Tx4a/=I 
40(a/  — l)2  = — 40.  Ans. 


2.  Multiply  a\/  — b by  c^/ — cl. 


a*fb  -V -1  x cV  cl  *a/— 1 

acVbd-(V  — l)' 
—ac*fbd.  Ans. 


3.  Multiply  (a  + b)V—{ci  + b)  by  (a  + b)V—(a  + b). 


Ans.  — ( a + b )3. 


70 


ELEMENTS  OF  ALGEBRA. 


5.  V-2-V-2  V-2  = -2vW-l. 

6.  («—  V — b)(ju  + a/~ b)=az  + b. 

7.  («+  V— #)(«  + V— b)=as  +2av^— 6 — 

8.  3a/5  x 5a/— 2=15^/— 10. 

9.  (3  + a/— 5)1-2—  a/—  5)  = 11  — V — 5. 

10.  a/ — «3  -a/ — £2  -V —c2  -a / d2-^/  —f2=abcdf. 

11.  (a/33  + V3i)(a/Z:12-a/^4:)=-1. 

12.  (a/—2  — a/3)(\/2 — V — ^)  = a/^— 9 + a/— 4- 

The  division  of  imaginary  quantities  of  the  second  degree  is  man- 
aged in  a corresponding  way,  being  careful  to  introduce  the  im- 
aginary factor  a/  — 1. 


Examples. 

1.  Divide  10a/  — 1 by  5 a/  — 1. 

2.  Divide  a a/  — by  ca/ — <7. 

3.  Divide  4a/ —a2b2,  by  2a/  — i3. 

4.  Divide  9a/  — 10  by  3a/ —2. 


Nns.  2. 

, «a/£ 

cVd 
Ans.  2a. 
Ans.  3 a/s'. 


5.  Divide  «2+5  by  u—  v7  — A A«s.  a + V—  A 

Imaginary  expressions  of  a higher  degree  than  the  second,  may 
be  treated  in  a corresponding  manner. 


SECTION  VI. 

THE  ROOTS  OF  NUMBERS. 

69.  The  extraction  of  the  roots  of  numbers  properly  belongs  to 
Arithmetic ; but  it  may  be  well  to  show  here  the  rationale  of  the 
process. 

Since  the  square  root  of  100  is  10,  we  know  that  the  square  root 
of  any  number  which  contains  but  two  digits,  must  be  expressed  by 


THE  ROOTS  OF  NUMBERS.  71 

a single  figure.  We  can  readily  find  the  root  of  such  a number  by 
trial,  when  it  is  a perfect  power. 

AYhen  the  number  contains  three  or  more  figures,  there  will  be  at 
least  two  figures  in  its  root ; that  is,  the  root  will  contain  a certain 
number  of  tens  and  a certain  number  of  units. 

Then,  let  n be  any  number,  and  a the  number  of  tens,  and  b the 
number  of  units  in  its  root.  The  square  root  of  n will  be  (a  + 5),  and 
we  may  write, 

n = {a  + b)~  =a~  + 2ab  + b2  • • • • (1). 

We  see  from  this  that  the  number  contains  a2,  that  is,  the  square 
of  the  tens,  and  2 ah,  that  is,  twice  the  product  of  the  tens  and  units  j 
and  b2,  the  square  of  the  units. 

Let  it  now  be  required  to  find  the  square  root  of  1441.  We  shall 
begin  by  finding  the  number  of  tens  in  the  root; 
and  since  the  square  of  a single  ten ' gives  units  1444i38 
followed  by  two  0’s;  thus,  (10) 2 = 100,  the  square  I 

of  any  number  of  tens  can  give  only  0’s  in  the  last  60|544 


two  places.  We  may  then  place  a point  over  the  480 

third  figure  from  the  right,  as  here  seen,  to  show  04 

that  the  last  two  places,  44,  may  be  regarded  as  oc-  _G4=  (8) 2 
cupied  by  0's,  in  our  search  for  the  tens  of  the  0 

root. 


Noav,  since  the  square  of  the  tens  of  the  root  must  be  found  ex- 
clusively in  14,  the  square  root  of  the  greatest  perfect  power  in  14, 
which  is  9,  will  be  the  tens  of  the  root.  3,  then,  is  the  first  figure  of 
the  root,  and  30  will  correspond  to  a in  formula  (1).  Write  the  3 to 
the  right,  as  shown.  Now,  squaring  it,  subtract  the  result  from  14, 
and  bring  down  the  44.  We  have  really  subtracted  (30) 2 = 900,  and 
the  remainder  is  what  is  left  after  taking  away  a 2 from  the  formula. 
It  must  then  correspond  to  the  2ab  + b2o‘L  the  formula.  We  should 
be  able  to  find  b at  once  from  this  remainder,  if  it  were  2ah  alone,  by 
dividing  by  2a,  (2  x 30),  a being  now  known.  But,  at  any  rate,  since 
b 2 is  quite  small  compared  with  2 ab,  we  shall  not  come  far  from  b by 
dividing  the  remainder,  as  it  stands,  by  2a.  Then,  doubling  30,  and 
dividing  the  remainder  by  the  result,  we  get  9.  Now,  multiplying  60 
by  9 (giving  2 ab),  and  subtracting  the  result  from  the  remainder,  we 
shall  have  4 left.  But  this  second  remainder  must  be  equal  to  the 
square  of  the  units,  b2  ; in  this  case  (9)2  = 81.  If  our  original  number, 
then,  is  a perfect  power,  9 is  too  great. 


72 


ELEMENTS  OF  ALGEBRA. 


Trying  8 we  find  64  fora  remainder,  and  tins  is  just  equal  to  (8) 2 : 
38,  then,  is  the  root  required. 

To  shorten  the  last  part  of  the  operation  we  may  double  the  3,  and 


divide  the  remainder,  exclusive  of  the  last  figure,  by  it ; 1444;38 

then  write  the  result  in  the  unit’s  place,  after  6.  Now,  

when  we  multiply  68  by  8,  we  form  at  once  (2 a + b)b  

=2  ab  + b2.  0 


When  the  number  requires  more  than  two  places  in  the  root,  we 
might  apply  the  same  reasoning  to  the  discovery  of  the  two  figures 
of  the  root  of  the  highest  denomination,  and  then,  regarding  all 
these  as  forming  one  denomination,  proceed  as  before.  This  is  ac- 


complished by  simply  pointing  off  from  the  right,  . . . 

in  places  of  two  figures  each,  and  continuing  the  15129  123 

operation  as  above.  If  we  had  two  more  places  of  1 

figures  in  the  above  example,  we  should  bring  them  22; 51 
down,  and  then  double  38,  and  divide  the  remain-  44 

der,  exclusive  of  the  right  hand  figures,  by  the  re-  243]  729 
suit,  writing  it  in  the  root,  and  also  in  the  new  trial  729 

division,  and  so  proceed  as  before.  0 


70.  Extraction  of  the  ?fth  Hoot  of  Numbers. 

Let  it  be  required  to  find  any  root  of  a number,  JV.  The  root  will 
consist  of  a tens  and  b units.  Thus  we  shall  have, 

N—(a  + l)"=an  + nan-'b  + , etc. 

Now  the  wtli  power  of  the  tens  must  have  O's  in  the  last  n places, 
so  that  in  looking  for  the  tens  of  the  root  we  may  point  off  the  last 
n places  of  the  number.  We  may  now  find  the  tens  of  the  r jot  by 
finding  the  highest  perfect  root  in  the  left  hand  period.  Subtracting 
the  «th  power  of  the  tens  so  found  from  the  number,  we  shall  have 
na"~lb  + , etc.,  left.  Now,  by  forming  n times  the  «■  — 1 power  of 
the  tens,  and  dividing  the  remainder  from  it,  we  shall  have  the  units 
of  the  root,  or  something  too  great.  By  raising  the  trial  root  so 
found  to  the  «th  power,  we  shall  find  whether  it  produces  the  num- 
ber or  not.  If  too  great,  the  figure  in  the  unit’s  place  must  be 
diminished. 

We  shall  now  give  an  example  in  cube  root.  Any  other  root  may 
be  found  m like  manner. 


THE  BOOTS  OF  NUMBERS. 


73 


Let  us  find  the  cube  root  of  12977875. 

It  will  be  observed  12977875  [235 

8 

that  only  the  first  fig-  o 22  _ , 0175 


ure  of  the  second  period  12977  = 1st  two  periods. 

(23)3  = 12167 

is  brought  down  for  a 3 x (23)  * = ISBTjsIoS 


first  remainder.  This  12977875  = The  three  periods. 

. , (235)3  = 12977875 

is  because  12,  the  first  q 

divisor,  is  really  1200,  so  that  the  remaining  two  figures  would  be 
cut  off  in  dividing,  if  brought  down. 


Examples. 


1.  Find  the  square  root  of  651219.  Ans.  807. 

2.  Find  the  cube  root  of  12167.  Ans.  23. 

3.  Find  the  cube  root  of  421875.  Ans.  75. 

4.  Find  the  square  root  of  1058841.  Ans.  1029. 

5.  Find  the  cube  root  of  256047875.  Ans.  708. 

6.  Find  the  fifth  root  of  248832.  Ans.  12. 


Remark.  — Let  it  be  remembered  that 


\ / V/«  = v/«,  and  that,  therefore. 


when  the  index  is  a multiple  of  two  or  more  factors,  we  may  find  the  roots  in- 
dicated by  those  factors,  successively,  instead  of  extracting  the  entire  root  at 
once. 


7.  Find  the  sixth  root  of  244141625. 


Ans.  25. 


8.  Find  the  eighth  root  of  214358881.  Ans.  11. 

Remark. — If  the  number  is  partly  or  altogether  decimal,  begin  to  point  off 
from  the  decimal  point  going  to  the  left  for  the  entire  part  and  to  the  right  for 
the  decimal.  Any  number  of  0’s  may  be  added  to  the  decimal  and  the  opera- 
tion thus  continued  to  any  degree  of  approximation. 

9.  Find  the  square  root  of  657.4096.  Ans.  25.64. 

10.  Find  the  square  root  of  .140625.  Ans.  .375. 

Remark. — A common  fraction  may  be  converted  into  a decimal  and  the  root 
extracted  to  any  desired  degree  of  accuracy. 

11.  Find  the  square  root  of  4j-9-. 


Ans.  2.3604. 


74 


ELEMENTS  OE  ALGEBBA. 


12.  Find  the  square  root  of  J-  to  five  places  of  decimals. 

Ans.  .74535. 

13.  Find  the  square  root  of  7,  to  four  places  of  decimals. 

Ans.  2.6457. 


71.  No  Exact  Boot  of  an  Imperfect  Power. 


The  a/5,  or  or  the  root  of  any  imperfect  power  of  the  degree 
indicated,  cauuot  be  found  exactly,  either  in  entire  or  fractional  quan- 
tities; hut  we  may  approximate  such  a root  as  nearly  as  we  please: 


for  example,  -v/7^2-64575131  + and  so  on  indefinitely. 

Such  quantities  are  said  to  be  incommeasurdble,  and  are  commonly 
called  surds. 

The  distinction  between  an  imaginary  quantity  and  a surd  is,  that 
we  may  obtain  the  root  of  a surd  as  nearly  as  we  please  without  being 
able  to  find  it  exactly,  while  we  cannot  make  the  first  movement 
towards  finding  the  root  of  an  imaginary  expression. 

Let  us  now  prove  that  we  cannot  find  the  exact  root  of  a surd. 

Let  p be  any  whole  number  whose  nth  root  is  | , a fraction  having 


no  common  factor  in  the  numerator  and  denominator.  ^ cannot,  then, 
be  reduced  to  an  entire  quantity.  TVe  shall  have, 


Raising  both  numbers  to  the  nth  power,  since  they  must  still  be 
equal,  we  shall  have 


a" 


Now,  in  raising  any  number  to  a power,  we  but  repeat  the  factors 
composing  it,  a certain  number  of  times,  introducing  no  new  ones, 
so  that  an  and  bn  are  still  prime,  with  respect  to  each  other ; that  is, 

(ln 

they  have  no  common  factor,  and  thus,  ^ is  an  irreducible  fraction  ; 


so  that  we  have  p,  a whole  number,  equal  to  an  irreducible  fraction, 
which  is  impossible.  Then,  since  y' p cannot  be  a whole  number 
nor  a fraction,  it  cannot  be  found  exactly  at  all. 


EQUATIONS. 


75 


SECTION  VII. 

EQUATIONS. 

72.  An  Equation  is  the  indicated  equality  of  two  algebraic  ex- 
pressions ; thus, 

ax+by=c+d 

is  an  equation. 

The  sign  = divides  the  equation  into  two  parts,  called  Members  : 
the  part  written  first  being  the  First  Member,  and  that  following 
the  sign  of  equality,  the  Second  Member. 

73.  The  essence  of  an  equation  (its  life,  so  to  speak)  lies  in  the 
equality  of  its  members.  Any  operation  may  be  performed  upon  it, 
so  long  as  the  equality  is  preserved  intact. 

The  following  are  self-evident  truths,  called  Axioms : 

1.  If  equal  quantities  be  added  to  both  members  of  an  equation, 
the  equality  will  still  subsist. 

2.  If  equal  quantities  be  subtracted  from  both  members,  the 
equality  will  still  subsist. 

3.  If  both  members  be  multiplied  by  the  same  quantity,  the 
equality  will  still  subsist. 

4.  If  both  members  be  divided  by  the  same  quantity,  the  equality 
will  still  subsist. 

5.  If  both  members  be  raised  to  the  same  power,  the  equality 
will  still  subsist. 

6.  If  the  same  root  of  both  members  be  extracted,  the  equality 
will  still  subsist. 

All  operations  upon  equations  are  founded  upon  these  axioms. 


74.  The  First  Transformation.  , 

There  are  four  principal  transformations  to  which  equations  are 
submitted.  We  shall  consider  them  m their  order. 

I.  The  object  of  the  First  Transformation  of  equations  is  to  clear 
an  equation  of  fractional  quantities. 

Take  the  equation, 


x 

a 


Now,  we  may  multiply  both  members  by  any  quantity,  and  still 


76 


ELEMENTS  OF  ALGEBRA. 


preserve  the  equality.  Let  us,  then,  multiply  each  term  by  aid,  the 
common  multiple  of  all  the  denominators.  We  shall  have, 


There  is  now  a common  factor  in  the  numerator  and  denominator 
of  each  fraction.  Stinking  them  out,  we  have, 


an  equation  in  which  all  the  terms  are  entire. 

It  is  generally  more  convenient  to  strike  out  the  common  factor 
from  the  multiple  before  multiplying  by  it.  We  may  thus  say,  that 
to  clear  an  equation  of  fractions,  or, 

To  make  the  first  transformation  of  an  equation,  form  the  least 
common  multiple  of  all  the  denominators;  divide  this  by  each  denom- 
inator in  succession  and  multiply  each  term  respectively  by  the  re- 
sult. Entire  terms  are  to  he  multiplied  by  the  common  multiple  as  it 
stands. 

It  is  sometimes  better  to  multiply  each  numerator  by  all  the  de- 
nominators except  its  own,  and  the  entire  terms  by  all  the  denom- 
inators. 


Clear  the  following  equations  of  fractions. 


Explanation—  Here  tlie  least  common  multiple  is  4b.  Dividing  it  by  2 and 
multiplying  the  numerator  of  the  first  term  by  2b,  we  have  2 bx  for  the  first  term 
of  the  result.  The  other  terms  are  found  in  like  manner. 


abd.x  aJid.u  alula 


bdx—ady=abc+ abdfi 


Examples. 


Ans.  2 to — 4a  + 3 be—  by — 4ib. 


Ans.  2a-x  + c—\Day  + 2ax. 


Ans.  3x—Za—x-+x=l2ax. 


1 _«  + £ 


Ans.  a—x—as+x2  = (a+x)2. 


a 


a2 


c 


Ans.  acVz—cV%—c=5a~b. 


EQUATIONS. 


77 


V x + y 


1 


a 


/Vx+y 


Ans.  x + y—aVx+y—a. 


Ans.  2b2x2 — a2y~1=12ab2. 


75.  The  Second  Transformation. 

IT.  The  object  of  the  Second  Transformation  of  equations  is  to  trans- 
pose terms  from  one  member  of  an  equation  to  the  other. 

Take  the  equation, 


How,  as  we  may  add  the  same  quantity  to  both  members  without 
affecting  the  equality,  let  us  add  — 2 by  to  both  members.  We  shall 
have, 


In  the  second  member  we  now  have  two  quantities  equal  with 
contrary  signs,  which  we  may  cancel,  and  we  shall  thus  have, 


It  will  be  observed  that  2 by  has  disappeared  from  the  second 
member,  and  appeared  in  the  first  with  its  sign  changed. 

How  suppose  we  desired  to  move  the  term  — cd  from  the  first 
member.  Adding  cd  to  both  members,  and  cancelling  as  before,  we 
have, 


The  term  has  disappeared  from  the  first  and  appeared  in  the 
second  with  a contrary  sign.  Thus, 

We  may  transpose  any  term  from  one  member  of  an  equation  to  the 
other  by  changing  its  sign. 

Examples. 

Transpose  all  terms  containing  an  unknown  quantity  to  the  first 
member  and  all  others  to  the  second. 


ax —cd—2by  + a. 


ax—cd—2by—2by  + a—2by. 


ax—cd—Tby—a. 


ax—2by=a  + cd. 


1.  ax—b—Zx2 — y. 

2.  a + 25  = — x2  +y. 


Ans.  ax—3x2+y=b. 
Ans.  x2—  y——  a— 25. 


ELEMENTS  OF  ALGEBRA. 


3.  21  a2b — xy—l—x. 

. a~  , x 

4.  — r 1 = ---4. 

o 2 


—xy—x=l—21a2b. 

A X A a~  i 

Ans.  - = 1 — - — 4. 
2 <y 


Re  mark. — Equations  are  usually  cleared  of  fractions  before  transposing,  but 
transpositions  may  be  made  at  any  time. 

Transpose  all  the  terms  of  the  following  to  the  first  member: 

5.  ax2—  1 = bx— .cxz.  Ans.  cxz  + ax2  — bx— 1=0. 

6.  0—a+jy—\y2  + yz—y/k.  Ans.  yi  — y3+\y2  — jy—a= 0. 

1)  /v  /v  y 


7.  *Ja  + b—  — {ci—b)x2  + V a-b-y. 


Ans.  ( a — b)x2  — Va—b-y  + Va  + b=0. 


8.  1: 


a + b 


a—b 


, a+b  x _. 

Ans.  r +1=0. 

x a — b 


Remark.— If  there  are  indicated  products  in  an  equation,  it  is  generally  better 
to  develop  such  expressions  before  transposing. 

In  general,  transpose  unknown  terms  to  the  first  member,  and  known  terms  to 
the  second. 

9 (a  + b)'2x  c__  (x—1)  x 3 . 

c 5 

5(a  + b)2x— 5r2  = (x—  1)  x 3c. 

5a2x  + 10abx  + 5b2x—5c2=3cx—3c. 
Ans.  5a2x  + 10abx  + 5b2x— 3cx=5c2  — 3c. 


76.  The  Third  Transformation. 

III.  The  object  of  the  Third  Transformation  of  equations  is  to 
so  change  an  equation  that  the  several  powers  of  the  same  unknown 
quantity  shall  enter  it  but  once. 

Let  us  take  an  equation  with  several  powers  of  the  same  unknown 
quantity ; thus, 

3ax2  — bx  + xz  — 2x  + x2  —axz  — 1. 

Writing  the  terms  containing  the  same  power  of  the  unknown  quan- 
tity together,  and  factoring  with  respect  to  these  several  powers,  we 
have, 

(l—a)xz  + (3a+l)x2—{b  + l)x2=l. 

Or  taking  a numerical  equation, 

5x2  — 2x2  +10.r— x=5  ; 


EQUATIONS. 


79 


Simplifying,  we  have, 

3x 2 +9x=5. 

If  an  equation  has  any  number  of  unknown  quantities,  the  same 
course  may  be  pursued. 

Practically  to  effect  the  third  transformation, 

Gather  together  the  terms  containing  the  like  'powers  of  the  same 
unknown  quantity  or  quantities,  and  factor  with  respect  to  those 
several  powers. 

Examples. 

Submit  the  following  to  the  Third  Transformation,  transposing 
unknown  terms  to  the  first  member,  and  known  to  the  second. 

1.  ax— l + cx2  — x + l=2xs.  Ans.  (c—2)x2  — (l—a)x=h—l. 

Caution. — Be  careful  in  putting  on  or  taking  off  a parenthesis  with  the  — 
sign  before  it. 

2.  l = a — V^-x2—  ax  + dx2  — x2  + ^fa-x. 

Ans.  (l  + V%  — d)x2  — (Va— a)x= a— 1. 

3.  3x2—  2x  + l=a2x2  —h3x  + cx+f. 

Ans.  (3—a2)x2  + (b3—c—2)x=f=l. 

t it  i i 

4.  d2x2  —cx—ax2  — 2x  + 25.  Ans.  (2— c)x+  (a2— a)x2—25. 

5.  3a;2  — 2x  + l — 7x2  — 4x.  Ans.  —4x2+2x—  — l. 

77.  Fourth  Transformation. 

IV.  The  object  of  the  Fourth  Transformation  of  equations  is  to 
make  the  co-efficient  of  the  highest  power  of  the  unknown  quantity 
unity. 

Let  us  take  an  equation  upon  which  the  first  three  transforma- 
tions have  been  already  performed ; thus, 

(3a— b)x2  — (4  + c)a;=l. 

We  may  divide  both  members,  that  is  to  say  every  term,  by  the  co- 
efficient of  x2,  and  we  shall  have, 

„ 4 + c 1 

— o ix=o I* 

3a — o 3 a — o 


80 


ELEMENTS  OF  ALGEBRA. 


The  same  course  may  be  pursued  in  any  case;  hence,  having  per- 
formed the  previous  transformations,  to  make  the  Fourth, 

J Divide  each  term  by  the  co-efficient  of  the  highest  power  of  the  un- 
known quantity. 


Examples. 

Apply  the  Fourth  Transformation  to  the  following: 
1.  (a  + b)x2  —x=b.  Ans.  x2 


a + b 


2.  {3  — c2)x3 — {a2 +b)x2 +2x—4:d3 —c. 

a2  +b 


Ans.  x3- 


3-c2 


x2  + 


3 — c2 


x = 


b 

a + b’ 

Ad3  — c 
3-c2  ’ 


3.  (2 —d2)x—a^—b2. 

4.  25x=40. 

5.  (a  — b)nx=cn. 


Ans.  x : 


2—d2 


Ans.  x = . 

2o 


Ans.  x ■ 


(a— b)m 


Perform  all  four  of  the  transformations  in  succession  on  the  fol- 
lowing; transposing  unknown  terms  to  the  first  member  and  known 
to  the  second. 


7. 


1. 


Ans.  x — 


be  + bed 
ac—b 2 


„ a2  + b2  c—d  , a 

8. 5-  X — X + T , 

c c~  b 


Ans.  x — 


ac2  +a2bc  + b3c 


bcl—bc—bc2 


10. 


25 

3 


5x  _ lO.r 

IT  ~ 1 ~ 12 


x. 


Ans.  x—— 102. 


Ans.  x — — 


88 

12  * 


78.  The  Change  of  Signs. 

Let  us  have  the  equation, 

— ax+by—  — c. 

Multiplying  both  members  by  —1,  we  have, 

ax—by=c, 


EQUATIONS. 


81 


an  equation  in  which  all  the  signs  have  undergone  a change.  Hence, 
we  may  change  the  signs  of  every  term  of  an  equation,  and  the  equality 
will  still  subsist. 

79.  Solution  of  Simple  Equations. 

An  equation  containing  but  the  first  poiver  of  the  unknown  quan- 
tity or  quantities  is  called  a simple  equation,  or  an  equation  of  the 
first  degree. 

It  will  be  observed,  that  when  we  have  submitted  a simple  equa- 
tion containing  a single  unknown  quantity  to  the  four  transforma- 
tions in  succession,  the  unknown  quantity  is  made  to  stand  alone  in 
the  first  member,  equal  to  known  quantities  in  the  second.  We 
have  thus  found  the  value  of  the  unknown  quantity,  and  are  said  to 
have  solved  the  equation* 

Then,  to  solve  a simple  equation  containing  but  one  unknown 
quantity,  we  have  but  to  submit  it  to  all  four  of  the  transformations 
in  succession.  These  four  steps,  in  few  words,  are, 

1.  Clear  the  equation  of  fractions  ; 

2.  Transpose  unlcnoion  terms  to  the  first  member  and  biown  to 
the  second  ; 

3.  Factor  the  first  member  with  respect  to  the  unlcnoion  quan  tity  ; 

4.  Divide  both  members  by  the  co-efficient  of  the  unknown  quan- 


The  value  of  the  unknown  quantity  so  found  is  called  the  Root  of 
the  equation.  The  root  of  an  equation,  then,  is  such  a value  of  the 
unknown  quantity  as  will  verify  the  equation  ; that  is,  when  substi- 
tuted for  that  quantity  in  the  equation  it  will  show  that  the  two 
members  are  identical ; thus,  the  root  of 


tity. 


is 


2 b 

2 a + b2’ 


Writing  this  in  the  equation,  we  have. 


Simplifying  and  clearing  of  fractions,  we  have, 
4 ab  + 2b3  = iab  + 2b3. 


F 


82 


ELEMENTS  OF  ALGEBRA. 


Thus,  the  two  members  are  entirely  the  same ; and  the  equation 
must  be  true. 

An  equation  may  be  made  to  have  any  two  equal  quantities  in  the 
two  members ; thus,  cancelling  equal  terms  in  both  members,  we 
have, 


The  verification  of  a numerical  equation  is  more  simple ; thus, 
solving 


we  have  x =50. 

This  for  x in  the  equation  gives, 


0 = 0. 


By  dividing  each  member  by  itself,  any  equation  will  be 


1 = 1. 


300-50=250 

250=250. 


Examples. 


Ans.  x=3. 


— 3x—12. 


Ans.  x=6. 


3*  4 6 2 3 


Ans.  x=3. 


Ans.  x= — 
a 


a + b 


5.  a”{x— 1)  +ab[x— 2)=b2. 


EQUATIONS. 


83 


ai~x  ~x  + 1=2. 


8. 


x 
c2 

7 + 9a; 


Ans.  x- 


5c$-2a$ 


Aiis.  x= 


35 

167' 


80.  Degree  of  Equations. 

The  degree  of  an  equation  is  determined  by  the  greatest  number 
of  times  the  unknown  quantity  or  quantities  enter  any  one  term  as  a 
factor;  thus, 

5x  1x—o  ) are  0£  £jje  grs£  degree> 

azx  + oi/—4:Z—c  ) ° 

\ are  of  the  second  degree. 
ci2xy  + ox=4:Z—c  ) ° 

5x3-7x=3  l 

„ , 7 . ■ are  ot  the  third  degree. 

a2xyz  + oxy=iz— c ) ° 


81.  Complete  and  Incomplete  Equations. 

A complete  equation  of  any  degree  is  one  which  contains  all  the 
several  powers  of  the  unknown  quantity  from  that  which  determines 
its  degree  down  to  the  zero  power;  thus, 

ax3  —bx2  + cx  + dx°  = 0 

is  a complete  equation  of  the  third  degree  of  one  unknown  quantity. 
x°  being  iinity,  we  may  write  it  or  not  as  we  please. 

When  one  or  more  of  the  intermediate  powers  are  wanting,  the 
equation  is  said  to  be  incomplete  ; thus, 

x 3 + bx2=d  'j 

x3  —5x=d  > are  incomplete. 
x3  — d / 


82.  Complete  Equations  of  the  Second  Degree. 

When  an  equation  has  undergone  the  four  transformations  already 
given,  it  is  said  to  have  its  Simplest  Form. 

Let  us  take  a complete  equation  of  the  second  degree  containing 
but  one  unknown  quantity ; thus, 


x2 

a 


x , 2x  , x2 

-2  + C = d~-3+T 


84 


ELEMENTS  OF  ALGEBRA. 


Submitting  it  to  the  four  transformations,  we  hare, 

. ab  ab(d-c) 

x +771 Vx=—7 "• 

6 (b—a)  b—a 

Now,  the  co-efficient  of  a;  is  a known  quantity,  and  we  may  let  2 \p  rep- 
resent it.  So  we  may  let  q be  equal  to  the  second  member.  Using 
these  values  in  the  equation,  we  have, 

x2 +2  px—q  - - (1). 

This  is  called  the  Reduced  Form  of  a complete  equation  of  the 
second  degree  containing  but  one  unknown  quantity.  The  terms, 
however,  may  have  different  signs. 

In  any  case,  after  making  the  sign  of  the  first  term  plus,  if  not 
so  already,  the  other  two  terms,  2px  and  q,  must  present  one  of  the 
four  following  phases : 

Both  plus; 

2px  minus  and  q plus; 

2px  plus  and  q minus; 

Both  minus. 

The  equation  (1)  written  in  every  possible  form  gives  the  follow- 
ing: 

x2  + 2px=  q. 
x2—2px~  q. 
x2  +2 px=  —q. 
x2—2 px——q. 

The  application  of  the  four  transformations  will  manifestly  bring 
any  equation  of  this  character  to  the  form  of  one  of  these  equations. 
They  are  thus  called  the  Four  Forms  of  a complete  equation  of  the 
second  degree. 

S3.  Solution  of  Incomplete  Equations. 

If  the  equation  is  incomplete,  there  will  be  no  term  containing  the 
first  power  of  the  unknown  quantity;  that  is,  2p— 0,  and  the  four 
forms  will  become, 

x2  — +q. 

We  may  extract  the  square  root  of  both  members  of  these  equa- 
tions, and  shall  have, 


x—±Vq. 


EQUATIONS. 


85 


It  will  be  observed  that  the  values  of  x in  the  last  case  are  ima°f- 
inary. 

We  have  thus  found  the  roots  of  the  incomplete  equations;  and  so 
in  any  case, 

To  solve  an  incomplete  equation  of  the  second  degree,  reduce  it  to  the 
form  of  x2=q,  and  then  extract  the  square  root  of  both  members. 

Reduce  the  following  equations  to  the  form  of  x2  + 2 px—q. 


1. 


a2x2 

~W 


2 ax  b2 

c c2 


:0. 


A ns.  x2 


2 ab2  c _ 54 

a2c2  1 a2c 2 


2. 


ce-l-12 

x 


X 

£ + 12 


26 

5 


Ans.  £2  + 12£— 45. 


3. 


ax  3x2  , 1 + 5 

T + -T  +1==  — i — 
o 4 b 


x 2 x 

4 + a' 


Ans.  x2  + 


a2  — b 1 
~aTX~b' 


Reduce  the  following  equations  to  the  form  of  x2=q,  and  solve. 

7i 

Ans.  x— 


4.  ax2  -\ — — — cx2  + 1. 
c 


c—b 


3x~  2 

5.  — =4.r3  — 14. 


ac  +c2 
Ans.  x=±2. 


7.  (a  + x)(a—x)  — ^~ 2x2. 


9.  ax(b— x)^=a[l  — 43)2. 


Ans.  x—  ±3. 


Ans.  x—  ± 


Ans.  x—  ±6. 


Ans.  x= 


84.  Solution  of  Complete  Equations. 

Resuming  the  equation, 

x2+2px=q  • • • • (1), 

we  see  that  the  first  member  may  be  made  a perfect  square  by  add- 
ing p2  to  it.  This  we  are  at  liberty  to  do,  provided  that  the  same 
quantity  be  added  to  the  second  member  also.  We  may  thus  have, 

x2  +2 px+p2=q+p2. 


86 


ELEMENTS  OF  ALGEBRA. 


Extracting  the  square  root  of  both  members,  we  have, 

x+p=±Vq+p2- 

Transposing  p to  the  second  member, 

x=~P±^qTp2 (2). 

These,  then,  are  the  values  of  x in  equation  (1).  There  are  always 
two  roots;  written  separately,  they  are, 

x=— p + ^q+p)2  and  x—  —p—  V q +JJ2- 

It  will  be  observed  that  the  entire  part  in  each  root,  —p,  is  alto- 
gether the  same,  and  is  half  the  co-efficient  of  the  first  power  of  x in 
the  equation  (1)  from  which  they  came,  taken  with  a contrary  sign  ; 
the  radical  parts  are  also  the  same,  but  one  of  them  has  the  plus, 
and  the  other  the  minus  sign. 

It  is  obvious  that  we  may  at  once  write  out  the  roots  of  such  an 
equation,  after  it  has  been  reduced  to  this  trinomial  form,  by  substi- 
tuting the  values  of  p and  q,  taken  from  the  equation  itself,  in  these 
roots. 

For  example,  let  us  have, 

ax2—-x=7—x~. 

k A 

The  application  of  the  four  transformations  gives  us, 

c _ 14 

r'"  2a  + 2‘X  ~2a+2' 

Q 

Now,  in  this  equation,  p,  the  half  co-efficient  of  x,  is  ^ - j-  ; and 
14 

* 13  2«T2- 

These,  in  the  formulas  for  the  roots  (2)  give  us 


c 

+ 4/  14  +1 

f c y 

4a + 4 

+/]/  2a  + 2 + 

c 

< / 14  1 1 

( c V3. 

4a + 4 

\ 2a  + 2 + 

Ua  + 4/  ' 

We  may  solve  any  such  equation  in  the  same  manner;  so  that  we 
can  say,  in  general, 

To  solve  a complete  equation  of  the  second  degree,  containing  one 
unknown  quantity, 


EQUATIONS. 


87 


1.  Apply  the  four  transformations  in  succession,  thus  reducing  it 
to  the  form  x2  -\-2pz=q. 

2.  Write  the  unknown  quantity  equal  to  half  the  co-efficient  of  the 
first  power  of  that  quantity,  with  a contrary  sign,  plus  and  minus 
the  square  root  of  the  second  member,  augmented  by  the  square  of 
this  half  co-efficient. 


Examples. 


1.  Solve 


10  14- 2x  22 


x x “ 


By  the  four  transformations, 


. 108  126 
^ 22  X~  22' 


Writing  out  the  roots, 

x=^±i/- 

22±y  22  +\22/  ' 

Performing  the  operations  under  the  radical  sign,  as  indicated, 

54  /Iir 

22±y  (22) 3 


Extracting  the  root. 


Whence, 


54  , 12 
^”“22  ± 22' 


21 


x=3  and  z=— . A ns • 


„ „ . 2 x2  bx  x x2  1 

2.  Solve \ — -2=z-r  + -~  + -■ 

a a2  b a a 


a2  —b2 


ab 


2=1. 


2 ab  ± V \ %ab  ) 


a2-b 2 , 
x—-  ^ -± 


2ab 


V 


ka2b2  +a2—  2a2b2  + bi 


4 a2b2 


ka2b2 


88 


ELEMENTS  OE  ALGEBRA. 


I A / ui  + Za*b2  +tA 
2ab  y Aci2b2 

a 2 — b2  a2  +b 2 

/y — 4- ; 

2a£  2ab 


2=7  and  2= — . Ans. 
b a 


3.  3x2 — 22=65. 

A 

s 

II 

Oi 

« 

ii 

i 

«£ 

4.  ax2—b=bx—^~. 

b2 

. b V4«33— 4a+J4 

Ans.  x=  — ± — — 

2 a 2ab 

5 * _ 7 
2 + 60  32—5 ' 

Ans.  2=14,  2=— 10. 

_ a2x2  2 ax  b2 

6.  T * + ^=°- 

0 A c cz 

b2 

Ans.  x—  — ±0. 
ac 

„ 48  _ 165  _ 

2 + 3 2 + 10 

A 27 

Ans.  2=—,  2=0. 
5 

0 2 + 12  2 26 

2 ' 2 + 12  5 

Ans.  2=3,  2=  —15. 

ax  3x2  , 1 +b  x2  x 
b 4 b 4 a 

. 1 a 

Ans.  2=-,  2=  — - . 
a'  b 

22—3  32—5  5 

U'  32-5  + 22-3  2 " 

i y 

Ans.  2=^  > 2=1. 

85.  Trinomial  Equations. 

A trinomial  equation  is  one  which  contains  but  two  different  pow- 
ers of  the  unknown  quantity.  Such  equations,  when  simplified, 
have  the  form, 

xm  + 2+2" = q. 

Complete  equations  of  the  second  degree,  such  as  we  have  just 
been  considering,  are  a particular  case  of  this  general  form,  m being 
2 and  n unity. 

The  method  just  explained,  of  solving  a complete  equation  of  the 
second  degree,  is  equally  applicable  for  the  solution  of  any  trinomial 
equation,  when  m—2n;  that  is,  when  the  equation  has  the  form, 

x"n  + 2+2" = q. 

We  can  make  the  first  member  a perfect  square  in  the  same  man- 


EQUATIONS.  89 

ner  as  in  the  last  article,  and  we  should  have,  after  extracting  the 
square  root  of  both  members  and  transposing, 

xn=  — p±Vq+pz. 

Now  extracting  the  nth  root  of  both  members,  we  have, 
x = y — p ± V q + p 2 ■ 

To  solve  such  an  equation,  then,  we  adopt  the  same  method  as  in 
the  case  of  an  equation  of  the  second  degree,  except  that  we  write 
the  unknown  quantity  with  an  exponent  equal  to  half  of  its  highest 
exponent  in  the  reduced  equation,  equal  to  the  roots,  and  after  that 
extract  the  nth  root  of  both  numbers  of  these  new  equations. 

For  example : 

a:4—  4:X2  — 12, 

whence,  x2  = 2 ± V12  + 4, 

and  thus,  x—  ± y2±'\/l6; ; 

or,  x—  ± ^/2±l. 

It  will  be  observed  that  there  are  four  roots  in  this  equation. 
Written  separately,  they  are, 

X—  + a/W,  x—  — t/fT,  x—  + a/— 2,  and  x=  — V — 2. 

It  will  be  observed,  further,  that  two  of  them  are  imaginary. 

It  may  be  well  to  remark  here,  that  every  such  equation  has  as 
many  roots  as  there  are  units  in  the  number  which  indicates  its  de- 
gree. Sometimes  some  of  the  roots  are  equal  to  each  other  and  some 
are  imaginary. 


Examples. 

1.  Solve,  x2*  — 2.t"  = 8.  Ans.  x—'sfi,  x—'\/—2. 

Remark. — Where  only  two  roots  are  given  in  tlie  answers,  the  student  may 
find  the  others. 

2.  x4  +x2  = 20.  Ans.  x=±2,  x=±V—5. 

3.  xG  +4x3  = 96.  Ans.  x=2,  x=  ■y/—l2. 

4.  (x  + y)2  — (x+y)  = (j.  Ans.  x-\-y— 3,  x + y=—  2. 

Remark. — Here  x+y  is  to  he  regarded  as  a single  quantity. 


90 


ELEMENTS  OF  ALGEBRA. 


5.  ( Z-1)2-(Z-1)  = 6 . 

6.  x—Vx—3. 


Ans.  x=4,  x=  — l. 
Ans.  Vx=3,  ^/x=—2. 
Ans.  's/x—Z,  /\fx=—2. 


7.  's/ x — \J x—G . 


86.  Simultaneous  Equations. 

Simultaneous  Equations  are  those  in  -which  any  quantity  in  one 
of  the  equations  is  the  same  in  quantity  and  quality  as  that  quan- 
tity in  any  other  of  the  equations;  thus,  if 


are  simultaneous  equations,  and  x in  the  one  has  a particular  value, 
as  5,  it  must  have  the  same  value  in  the  other;  and  again,  if  it  stand 
for  a particular  kind  of  quantity,  as  pounds,  in  the  one,  it  must  be 
pounds  in  the  other.  So,  also,  with  y,  or  any  other  unknown  quan- 
tities which  may  enter  such  equations. 

If  the  known  quantities  are  represented  by  letters,  the  same  thing 
is  true  of  them  as  well ; thus,  if 


are  simultaneous,  then  a must  have  the  same  numerical  value  and 
represent  the  same  kind  of  quantity  in  both  equations. 

It  is  plain  that  only  simultaneous  equations  can  be  combined,  as, 
for  example,  added  member  to  member,  or  multiplied  member  by 
member,  for  if  the  same  symbols  should  represent  different  things  in 
the  several  equations,  the  result  of  such  combination,  though  a true 
equality,  would  mean  nothing. 

Equations  are  commonly  combined  for  the  purpose  of  getting  rid 
of  one  or  more  of  the  unknown  quantities  which  enter  them ; or,  as 
it  is  said,  in  order  to  eliminate  such  unknown  quantities. 

There  are  three  methods,  commonly  in  use,  of  combining  equa- 
tions for  the  purpose  of  eliminating  unknown  quantities.  They  are, 


In  general,  the  first  thing  to  be  done  with  equations  preparatory 
to  combining  them  by  any  one  of  these  methods,  is  to  subject  them 
to  the  first  three  transformations.  The  same  unknown  quantity 


5x  + 3y= 10,  and 
3x  — 7y=lo, 


ax  + by—c 
dx—ay—b 


1st.  By  Addition. 

2d.  By  Substitution. 
3d.  By  Comparison. 


EQUATIONS. 


91 


will  then  enter  any  one  of  the  equations  but  once,  if  the  equations 
are  of  the  first  degree.  If  of  a higher  degree,  any  particular  power 
of  tlie  same  unknown  quantity  will  enter  but  once. 


87.  Elimination  by  Addition. 

Let  us  now  consider  these  three  methods  of  elimination  in  the 
order  given. 

Take  two  simultaneous  equations  of  the  first  degree,  one  numer- 
ical and  the  other  literal,  in  order  to  make  the  case  more  general, 
and  suppose  them  already  brought  to  the  proper  form ; thus, 

ox—oy—lQ 
ax+  by—c. 

How,  we  may  multiply  the  first  equation  through,  by  a,  and  the 
second  by  5,  without  disturbing  the  equality  in  either  case.  We 
shall  thus  have, 

5ax—3ay=10a. 

5ax  + 5by  —5  c. 

We  may  further  change  the  signs  of  all  the  terms  of  either  of  the 
equations,  and  still  preserve  the  equality.  Changing  the  signs  of 
the  second,  we  have,* 

5ax— 3ay=10a 

5ax  + 5by  — oc. 

How,  adding  the  equations  member  by  member,  we  have, 
—3ay—5by=—5c  + 10a. 

and  from  this, 

_5c  — 10a 
y 3 a + ob' 

In  like  manner,  we  may  find  the  value  of  x ; or,  we  may  substi- 
tute this  value  of  y in  either  one  of  the  equations,  and  find  it  in 
that  way;  thus,  writing  this  value  of  y in  the  first  of  the  given  equa- 
tions, we  have. 


* It  will  be  found  best  not  to  destroy  the  original  signs,  but  simply  to  write  the  new  ones 
below,  and  a little  to  the  right  of  the  old  ones,  as  here  shown. 


92 


ELEMENTS  OF  ALGEBKA. 


Solving, 

_106  + 3 c 
X 3«  + 56‘ 

It  will  be  observed  that  the  object  in  this  method  of  elimination 
is  to  make  the  terms  containing  the  particular  unknown  quantity 
which  we  want  to  be  rid  of,  cancel  when  we  come  to  add  the  equa- 
tions. These  terms  must  be  made  entirely  the  same  and  must  have 
different  signs,  in  order  to  do  this.  We  multiply  or  divide  the  two 
equations  by  any  quantities  which  will  cause  the  terms  in  question 
to  become  alike,  and  if  they  have  not  already  different  signs,  we 
change  all  the  signs  of  one  of  the  equations  to  make  them  differ. 

We  may  say,  then,  practically, 

To  eliminate  a quantity  by  addition, 

1.  Subject  the  equations,  if  necessary,  to  the  first  three  transform- 
ations ; 

2.  Multiply  or  divide  either  or  both  equations  by  whatever  will 
rnalce  the  terms  to  be  cancelled  the  same  ; 

3.  Make  the  signs  of  these  terms  differ,  if  they  do  not  already,  by 
changing  all  the  signs  in  one  of  the  equations,  and  then  add  member 
to  member. 

The  resulting  equation  will  be  free  from  the  quantity  in  question. 

Examples. 

..  c , 4x  _ x+2  y 

2 4 

-v  „ 24  — 6.r 

Simplifying,  we  have, 

”tx-  2y=20 
4Sx  + 24y=2L 

Dividing  the  second  equation  by  12, 

qx-2y=20 
4x  + 2y—2. 

Adding  and  then  solving, 

x=2. 

This  value  of  x in  any  of  the  above  equations  gives, 

y=~3- 


EQUATIONS. 


93 


2.  Solve 


3.  Solve 


i3x-\-3y—21 
\iy —2x—8. 

!ax=by 
x—c—y. 


Ans.  x—2,  y— 3. 


, be  ac 

Ans.  x — — — r , y — - 


(5s-8i=7y— 44 
4.  Solve  \ J 

(2  x=y+%. 


5.  Solve  ■< 


!ax  + by=c 
fx+gy-h. 


a + b’ J n + b' 
Ans.  x=4£,  y= 8f. 


. cq—bh  ali—cf 

Ans.  x — — — — , y = -f-,. 

ag-bf  J ag-bf 


6.  Solved 


1.3^72,-341=  If + ?|r 


2z+f=l. 


a 1018  W135 

Ans.  *=-18g^,y=-71^. 


7 Solye  |(*+5)(y+7)=(*+i)(y-9)+ii2 

( 2*  + 10=3y  + l.  x=3,  y=5. 


(sx+o  y=  <j=aal 

I ^ bz—az 


8.  Solve  < 


a2  be 


bzx — - + («  + b + c)ay=azx(2a  + b)ab. 


. ab  ab 

Ans.  x—  , yz 


b—a ’ J a + b" 


88.  Elimination  by  Substitution. 

Resuming  the  equations  in  the  last  article, 

6z-3y=10, 

cix+by—c, 

let  us  find  the  value  of  one  of  the  unknown  quantities,  as  x,  in  terms 
of  the  other  in,  say,  the  first  equation  ; that  is,  let  us  regard  all  the 
quantities  in  the  first  equation  as  known,  except  x,  and  then  find  the 
value  of  x,  under  this  hypothesis.  We  shall  have, 

104-3?/ 
x— 


5 


94 


ELEMENTS  OF  ALGEBRA. 


Now,  we  may  place  this  value  ol'  x for  that  quantity  in  the  second 
equation ; thus, 

o(10+%)+Jy=a 

We  have  now  an  equation  with  but  one  unknown  quantity  in  it. 

Solving  this,  we  have, 

^ __5c— 10a 
^ 3a  + hb‘ 

Substituting  this  result  for  y in  either  of  the  equations,  we  have, 

,_10£  + 3c 
da  + hb' 

This  is  the  method  of  elimination  by  Substitution.  We  may  thus 
say  that, 

To  eliminate  a quantity  by  Substitution,  find  the  value  of  such  quan- 
tity in  terms  of  the  remaining  quantities  from  one  of  the  equations, 
and  substitute  this  value  in  the  other. 

The  resulting  equation  will  be  free  from  the  quantity  in  question. 


Examples. 


1.  Solve < 


*_i-2  _L 
2 "3 

2z=^  + 6. 

O 


j'  x—y— 1 

2.  Solve  ■{  y 

x—~—2x—ll. 

I 6 


3.  Solve 


4.  Solve 


x+y—a 

x—y—b. 

2x  3 y __  9 

F+T_20 

3x  2 y _ 61 

T + y_i20< 


Ans.  x—4,  y= 3. 


Ans.  x=8,  y—  9. 


a + b 

T 


Ans.  x=—r~  , y- 


a—b 


. 1 1 

Ans.  x—-^,  y—-. 

/V  O 


5.  Solve" 


x _ v 

- + 5=fr-® 

a 2 


V , 

x+f~i —ay. 


Ans.  x—  - - - , y- 


Let  the  examples  in  the  previous  article  be  solved  by  this  method. 


EQUATIONS. 


95 


89.  Elimination  by  Comparison. 


Take,  again,  the  equations 

5x—Zy=lQ 
ax  + by—c. 

Find  the  value  of  x in  both  equations  in  terms  of  y ; thus, 

10  + 3?/ 
x= — - — -. 
o 


-by 


Tliese  must  be  equal  to  each  other,  and  thus  we  have, 
10  + 3?/  _ c— by 


Solving,  we  have,  as  before, 

_5c—10a 
y 3a + 56’ 

Finding  x in  the  same  ivay  or  by  substitution,  we  have, 

_106  + 3c 
X 3a  + 56’ 

This  is  called  the  method  of  elimination  by  Comparison.  Prac- 
tically, 

To  eliminate  a quantity  by  Comparison,  find  tlic  value  of  the  quan- 
tity in  both  equations  in  terms  of  the  remaining  quantities,  and 
place  these  values  equal  to  each  other. 

The  resulting  equation  will  be  free  from  the  quantity  in  question. 


Examples. 


1.  Solve 


16z + 30?/= 18 
12a;— 2+?/=  —2. 


;=99— 7 y 


7®=51-|. 


Ans.  z=|-,  y=\. 


2.  Solve  < 


Ans.  x=H,  y= 14. 


96 


ELEMENTS  OF  ALGEBRA. 


3.  Solve 


ax  + by=c 
dx+fy—g. 


4.  Solve 


j"  abx— | 
\^aby—l—x. 


Ans.  x 


fc—  bg  _cd—ag 
fa—bd y bd—af 


Ans. 

U 


1 _ 2 ab 

2aW+l'V~  2a2b*- 1' 


5.  Solve  -j  * + ?/=18.73 

( 0.56a;  + 13.42?/= 763.4.  Ans.  x—  — 39.81 

y—  58.54. 

Let  the  examples  in  the  last  two  articles  he  used  for  practice  by 
this  method. 


90.  Elimination  in  General. 

Let  us  now  extend  the  principle  of  elimination  to  the  solution  of 
equations  containing  any  number  of  unknown  quantities. 

Let  us  have  three  equations  which  have  been  already  subjected  to 
the  first  three  transformations,  and  containing  three  unknown  quan- 
tities; thus, 

2^  + 3 y—  z=l  - - - - (1). 

"5x— 2y  + oz=2  - - - - (2). 

x+  y—  2=3  - - - - (3). 

Manifestly  we  may  eliminate  one  of  the  unknown  quantities  from 
any  two  of  these  equations,  by  either  of  the  three  methods  already 
given.  Let  us,  say,  combine  the  first  two,  and  get  rid  of  2.  Multi- 
plying the  first  by  3,  and  adding,  we  have 

§x  + 1y=o  - - - - (4). 

In  the  same  way,  combining  the  first  and  last,  eliminating  z,  we 
have 

x+2y=— 2 - - - - (5). 

We  now  have  two  equations  (4)  and  (5),  with  hut  two  unknown 
quantities.  These  we  may  now  combine  by  any  one  of  the  three 
methods,  and  shall  have 

24  , 23 

x=  - and  y--Tv 


EQUATIONS. 


97 


Substituting  these  in  either  of  the  three  original  equations,  we 
have 


2 = 


32 

11' 


We  have  thus  solved  the  group  of  equations.  The  same  course 
may  be  pursued  with  any  number  of  equations,  provided  that 
there  are  as  many  equations  as  there  are  unknown  quantities. 

Thus  we  may  say,  that 

To  solve  a group  of  simultaneous  equations,  there  being  as  many 
equations  as  there  are  unknown  quantities, 


Combine  the  equations  two  and  two,  being  careful  not  toxmite  the 
same  two  more  than  once,  and  always  eliminating  the  same  quantity, 
until  there  are  as  many  resulting  equations  as  there  are  unknown 
quantities  remaining. 

Combine  these  resulting  equations  in  the  same  way.  We  shall  at 
last  have  an  equation  containing  a single  unknown  quantity,  which 
may  be  solved.  Complete  the  solution  by  substituting  the  value  of  the 
quantity  so  found  in  one  of  the  resulting  equations , with  two  unknown 
quantities  in  it,  and  so  on. 


Examples. 


r x+y— 10 

1.  Solve  •<  x+  2=19 
i y + 2=23. 


At 

y=±1-^ 

A 

41  2 

2.  Solve  { x_~2  4 


5- 


( x+y+z= 30 

3.  Solve  -j  8.3  + 4?/  + 22=50 
( 27.3  + 9y  + 3z= 64. 


Ans.  x—3,  y— 7,  2=10. 


Ans.  3=18,  y= 32,  2=10. 
Ans.  3=f,  y=—  7,  z=36£. 


4.  Solve  < 


x+y=a—z 

by—cx 

dz—fx. 


Ans.  x 


abd 

bd  + cd  + bf’ 


acd 

V~  bd+cd+bf’ 


_ a^f 

bd  + cd +bf 


G 


98 


ELEMENTS  OP  ALGEBRA. 


(1,1 
- + -=« 

2 

A?is.  x — 

* y 

a + b-c ’ 

5.  Solve < 

-+\=b 

2 

x 2 

^ a—b  + c’ 

1 1 
— h - ~C. 

2 

z — 

Ly  z 

b + c—d 

91.  Equations  of  Condition. 

If  we  should  have  more  unknown  quantities  than  we  have  inde- 
pendent equations,  wre  shall  have  more  than  one  unknown  quantity 
in  the  last  resulting  equation. 

If  we  should  have  more  equations  than  there  are  unknown  quan- 
tities, we  may  eliminate  all  the  unknown  quantities,  and  thus  have 
a resulting  equation  between  known  quantities;  thus,  let  us  have, 

y=ax+b. 
y—cx  + d. 

y =/%+!/■ 

Placing  the  second  members  of  the  first  two  equal  to  each  other, 
we  have, 

,7  ,7  i 

ax  + b—cx  + d :.x— . 

a—c 

Then  combining  the  first  equation  with  the  third,  eliminating  y,  we 
have, 

ax  + b=fx+y  .*. 

9~b 
A «-/' 

Placing  these  two  values  of  x equal  to  each  other,  we  have 
^ — y>  an  equation  in  which  there  is  no  unknown  quantity. 

Now,  if  this  last  result  is  not  a true  equality,  all  the  equations 
from  which  it  was  derived  cannot  be  true.  They  could  not  all  be 
satisfied  at  the  same  time  for  the  same  set  of  values  for  the  unknown 
quantities.  Such  a resulting  equation  between  constants  is  called  an 
Equation  of  Condition;  since  it  is  the  condition  upon  which  the  equa- 
tion from  which  it  is  derived  can  be  true. 

When  there  are  more  equations  than  there  are  unknown  quanti- 
ties, there  must  thus  be  a certain  interdependence  between  the  con- 
stants which  enter  them.  The  equations  from  which  such  equation 
of  condition  is  derived,  cannot,  therefore,  be  independent  equations. 


EQUATIONS. 


99 


Examples. 


Find  the  equation  of  condition  in  the  following  groups  of  equa- 
tions : 


r ax=b 
l cx=d 


2.  ^ 


ax— by 
cy=dx 


l fx=g • 


3. 


j y—ax  + b 
( y=a'x  + b. 


Ans. 


t_d 
a~  c 


Ans. 


cJl_ag 
¥ ¥ ' 


Ans.  a=a'. 


92.  Simultaneous  Equations  of  a Higher  Degree. 

Equations  of  a higher  degree  than  the  first  can  be  combined  and 
a single  equation  found  containing  a single  unknown  quantity.  The 
method  of  elimination  is  altogether  the  same  as  that  already  ex- 
plained ; but  the  resulting  equation  is  generally  of  too  high  a degree 
to  be  managed  by  methods  falling  within  the  province  of  this  work : 

For  example,  let  us  have, 

ax2  + by = c. 

2y2-3x=5. 

Solving  the  first  equation  with  respect  to  y,  we  have, 

c—cix 2 


This  substituted  in  the  second,  gives, 

0/h— «z2\2 
\—r-)  -3z=o, 

which  is  an  equation  of  the  fourth  degree ; and,  in  general,  when 
both  equations  are  of  the  second  degree,  the  resulting  equation  will 
be  of  the  fourth. 


93.  Simultaneous  Equations  of  the  First  and  Second  Degrees. 

There  are,  however,  two  classes  of  simultaneous  equations,  beyond 
the  first  degree,  which  admit  of  ready  solution.  They  are, 

1.  Equations  containing  twkunhnown  quantities,  when  one  is  of  the 
second  degree  and  the  other  of  the  first. 

2.  When  the  equations  are  both  of  the  second  degree  and  homogeneous 
with  respect  to  the  unknown  quantities. 


100 


ELEMENTS  OP  ALGEBRA. 


I.  Let  us  take  the  first  case.  Assume, 

Gx2  — 2xy=36 
3x  + y= 12. 

The  value  of  y,  in  the  second,  in  terms  of  x,  is, 

y=12— 3x. 

This  substituted  in  the  first  gives, 

6a;2  — 2a;(12— 3x)=36 ; 
whence,  x2—2x=3. 

x—3  and  —1. 

These  values  in  the  second  equation  give, 
y— 3 and  15. 

A like  course  may  be  pursued  in  any  such  case,  so  that  we  may 
always  solve  two  simultaneous  equations  containing  two  unknown 
quantities,  when  one  of  them  is  of  the  first  and  the  other  of  the 
second  degree,  by  simply  eliminating  one  of  the  unknown  quan- 
tities. 


94.  Homogeneous  Equations  of  the  Second  Degree. 

II.  When  the  equations  are  homogeneous  Avith  respect  to  the  un- 
known quantities. 

Take  the  equations, 

x2—2xy=l—3xy  - - (1), 

y2  + 4a;2  — 2 = 5a;2  - - (2). 

The  solution  is  accomplished  by  using  an  auxiliary  quantity. 
Applying  the  second  and  third  transformations  and  making  y=2}X’ 
Ave  have, 

x2  +px2  = l.  .-.  **  = — - - (3), 

p2x2  — x2 =2.  .*.  x2  — t - - (4). 

p~  — l 

Placing  these  tivo  values  of  x2  equal  to  each  other, 

1 _ 2 
l+p—p2—l' 
p2  — 2p—3 

p-i±V3+i 
p>=3  and  —1. 


Whence. 


EQUATIONS. 


101 


Using  the  first  value  of  p in  either  (3)  or  (4),  we  have, 

x=  ±-i. 

The  first  values  of  ^ and  x in  y=px,  give, 

y=b 

The  same  course  may  be  pursued  with  any  two  equations  of  this 
class. 

Examples. 


1.  Solve  x 2 + y2  = 13 


x + y— 5. 

CO 

II 

CO 

II 

2.  Solve  x2  +ys  =41 

x-y=-l. 

Ans.  x=4,  y=  5. 

3.  Solve  x2+y2  = 13 

%y-y2=2- 

Ans.  x=3,  y = 2. 

4.  Solve  x2  + 2xy+y2=2o  $ 

0 , x 1 

3x—4y=-  + j. 
J 4 4 

Ans.  x=  3,  y—2. 

5.  Solve  3x2-2y2=57 

3x  + y 
2 

Ans.  x=5,  y= 3. 

6.  Solve  3xy—2x2  = 10 

1 1 T’S  ,;/2 

2xy  + 2y2-  7 ■'  +25. 

Ans.  x=2,  y= 3, 

7.  Solve  x+y— 13 

Vx  + \/y=5. 

Ans.  x=9,  y= 4. 

8.  Solve  ‘\fxy  + x= 15 

V'x—  \/9  = l. 

Ans.  x=9,  y= 4. 

95.  Solution  of  Simultaneous  Equations  of  any  degree. 

The  principles  already  established  will  enable  us  to  solve  many 
equations  of  the  second  and  higher  degrees  by  the  exercise  of  a little 
ingenuity.  No  specific  rules  would  prove  of  much  service;  so  that 
the  general  management  of  such  cases  may  be  best  exemplified  by  a 
few  examples. 

1.  Solve  x3—y3  = 19  - - (1), 

x -y  = l - - (2). 


102 


ELEMENTS  OF  ALGEBRA. 


We  may  divide  the  first  equation  by  the  second,  and  thus  have, 
x~+xy  + y~  — 19  - - (3). 

Now  the  value  of  x from  the  second,  a;=l  +y,  in  equation  (3), 
gives, 

y^+y-6. 

Whence,  y=—\+zy/  6+£ 

y=2,  y=  — 3. 

These  values  of  y in  (2)  give, 

x—  — 1,  x——2. 

2.  Solve  (x  + y)2  + (x  + y)  = 30  - - (1), 

x2-y2=o  - - (2). 

From  (1)  we  have, 

x+y—  — 1-±  V30  + 1, 
x+y——\±ii~,  x + y=5,  x + y=—6. 

Dividing  (2)  by  these  values,  we  have, 

x—ry—1,  x—y=  — f. 

From  these  values  of  x + y and  x—y,  we  have, 
x=3,  x——\. 

y= 2,  y=V—V-- 

3.  Solve  x3—y3  =11T  - - (1)> 

(x—y)2  — {x—y)  = (j  - - (2). 

From  (2)  we  have 

x—y— 3,  x—y—  —2. 

Dividing  (1)  by  this  first  value  of  x — y we  have, 
x2+xy  + \y2  — 39  - - (3). 

Substituting  the  value  of  x from  x—y—o,  in  (3)  and  solving,  we 
have, 

*/=2,  y=-5, 

£c=5,  x—  —2. 

4.  Solve  x3  + y3  = 152  - - (1), 

x +y  =8  - - - (2). 

Dividing  (1)  by  (2),  we  have, 

x2—xy  + y2  = 19. 


EQUATIONS. 


103 


Combining  this  with.  (2),  we  have, 

x=5,  x=3. 

Whence,  y =3,  y= 5. 

5.  Solve  x2+y2  = 74:  - - (1), 

xy  =35  - - (2). 

Multiplying  (2)  by  2,  wre  have, 

2xy=70  - - (3). 

Adding  this  to  (1),  we  have, 

x 2 +2xy+y2  = 144. 

Whence,  x+y=  12  - - (4). 

This  with  (3)  will  complete  the  solution ; or,  subtracting  (3)  from 
(1),  we  have, 

x2  — 2xy  + y2=4:. 

Whence,  x—y=2. 

This  with  (4)  gives, 

x=7,  y= 5. 


6.  Solve  x— y + \/x— y=3 
x+y= 8. 


Follow  the  same  general  course  as  in  the  second  example. 

Ans.  x=6,  y= 2. 

7.  Solve  x2+yz  = 9 - - - (1), 
x2y  + xy2  = 3 - - (2). 

Multiplying  (2)  by  (3)  and  adding  to  (1),  we  have, 
x3  + 3x2y  + 3xy2  +y3=  27. 

Taking  the  cube  root  of  this, 

x + y=3  - - (3). 

We  may  write  (2)  thus, 

xy{x  + y)  = 3. 


This  with  (3)  gives, 

3 xy=3  - - (4). 

From  (3)  and  (4)  we  have 

x=l,  y= 2. 


8.  Solve  Sx3  +128y3  =3520 
2a:  + 4y=16. 


Ans.  x=2,  y= 3. 


104 


ELEMENTS  OF  ALGEBRA, 


9.  Solve  x + y— 10 
f\fxy= 4. 


Ans.  z=8,  y— 2. 


10.  Solve  -*-1=3 

y 2 


Ans.  2=6,  y— 3. 


11.  Solve  x2  +2xy  + _?y2=64  . 

x2—y2  = — 16. 

12.  Solve  re2  — 3x  + xy=li  + 3y 

x 2 + xy  + 52=70 — 5 y. 

13.  Solve  3xy-\-Ax— ty 2 +10=2?/ 

1 5x2  — 62  + iy  + 3 = lOxy. 

14.  Solve  x+^/xy  + ?/=14 

a:2  +xy+y2=  84. 


2=3,  ?/= 5. 


Ans.  2=5,  y= 2. 


Ans.  x=l,  y—2. 


Ans.  2=2,  y— 8. 


96.  Radical  Equations. 

We  shall  now  give  some  further  examples  of  the  manner  of  solving 
equations  containing  radical  quantities.  No  invariable  method  can 
he  pointed  out. 

If  there  is  but  one  radical  in  the  equation,  we  may  make  it  stand 
alone  in  one  member,  and  then  by  raising  both  members  to  a power 
equal  to  the  index  of  the  radical,  it  will  disappear.  If  there  are  two 
radicals,  by  placing  them  in  different  members  and  raising  to  two 
successive  powers,  they  may  be  made  to  disappear;  but  generally  the 
resulting  equation  is  of  too  high  a degree  to  be  easily  managed. 

When  there  are  fractional  quantities  entering  the  equation,  it  may 
often  be  greatly  simplified  by  suppressing  common  factors. 


1.  Solve  5 — Vz  + i=x. 
Transposing, 


\/x  + i=x—5. 


Squarin 


2 + l=22  — 10a: + 25. 


Solving, 


2 = S,  2 = 3. 


2.  Solve  ~ — l = '\/2x—i. 


V 2—1 


EQUATIONS. 


105 


Dividing  the  numerator  of  the  fraction  by  the  denominator, 
V^  + l — 1 — V^x—4: 

Vx—V'^x—i. 

Squaring, 

x=2x— 4 

X—4:. 


3.  Solve  a + Vx—b—Vx  + a2. 

Squaring,  a2  +2aVx—b  + x—b^=x+a2. 

Simplifying,  2aVx—b=b. 

Squaring  again,  4a 2 (a;— b)  = b2. 

b 2 +4  a2b 


Whence, 

4.  Solve 
Squaring, 
Simplifying, 


* ‘ 4a2 

a Vx  + a_Va—x 
x b ~ b 

a 


2 V ax  + a2  x + a_a—x 

x bVx  + b2  ~ b2  ' 


a 


2 Vax  + a2 

bVx 


= 0. 


Transposing  and  squaring, 


4ax  + 4as 
b2x 


Clearing  of  fractions,  a2b2  =4 ax2  +4 a2x; 


whence, 


„ ah2 

x2  +cix=——. 


5.  Solve 


a , 

■T=-j± 


ab2  a2 

~T  + T> 


—a±V  ab 2 +ct 2 
*= 1 * 

x—a  -Vx  + Vn  r- 
+ — V x. 


V x—  V a 


x—a 


Getting  rid  of  common  factors, 


Vx  + V a + 


Vx—V  i 


:—\/x. 


106 


ELEMENTS  OF  ALGEBRA. 


Clearing  of  fractions, 

x—a  + l=x—  '/ax. 
Simplifying,  and  transposing, 

'/ax—a  — l. 

Squaring  and  dividing  £>y  a,  x=  ^ — 

6.  Solve  '/x—  / x— 5 = 1. 


7.  Solve  a—b=Vx—2ab. 

1 Vx  + 1 


8.  Solve 


's/x—l  a 
9.  Solve  VYx+l  — 2 — //+ll. 

Vx  + o 


/x  + l. 


10.  Solve 


x—9 


-—/x  — o. 


Ans.  a; =9. 
Ans.  x=a2  + i2. 

2a + 1 


Ans.  x- 


a- f 1 
Ans.  x=o. 

Ans.  x =4. 


11.  Solve 

12.  Solve  1 + Vl  + x=2. 

13.  Solve  |'/ a + /x—c. 

14.  Solve  x2  + (x— 9)^=1 — - — . 

(x—9)2 

15.  Solve  |/l6  — Vx2  + 45=^. 

16.  Solve  -+  (aS 

X X 0 


Ans.  x—a3c—i. 
Ans.  a: =8. 
Ans.  x—(cm—a)”. 
Ans.  x=12. 

Ans.  x=2. 
Ans.  x—{2 ai—i2)~. 


97.  Inequalities. 

We  shall  now  give  a few  principles  witli  regard  to  Inequalities. 

Two  inequalities  are  said  to  subsist  in  the  same  sense  when  the 
greater  quantity  is  in  the  same  member  in  both;  thus,  5>2  and 
7>3  are  in  the  same  sense.  When  the  greater  quantity  is  in  the 
first  member  of  one,  and  in  the  second  of  the  other,  they  subsist  m 
a contrary  sense;  thus,  5>2  and  3 <7,  are  in  a contrary  sense. 


EQUATIONS. 


107 


1.  We  may  add  the  same  quantity  to  loth  members  of  an  inequal- 
ity, or  subtract  the  same  quantity  from  both  members,  without  chang- 
ing the  sense. 

For  example,  add  5 to  both  members,  and  then  subtract  it  from 
both  members  of  the  following  : 

7>3. 

We  have, 

12>8,  and  2>— 2. 

If  we  subtract  10,  we  hare, 

— 3>  — 7. 

This  last  result  is  true,  considered  in  an  algebraic  sense. 

This  principle  enables  us  to  transpose  terms  from  one  member  of 
an  inequality  to  the  other,  by  changing  the  signs,  as  in  an  equation. 

2.  If  both  members  of  an  inequality  be  multiplied  or  divided  by  the 
same  positive  quantity,  the  sense  will  not  be  changed ; but  if  the  mul- 
tiplier be  negative,  the  sense  will  be  changed. 

For  example,  multiply  both  members  of  7> 5 by  5.  We  have, 

35>25. 

Multiplying  by  —5,  we  have, 

-35  <-25. 

If  we  multiply  by  —1,  we  change  the  sense  and  the  signs  at  the 
same  time ; hence,  we  may  change  the  signs  of  an  inequality  if  we 
at  the  same  time  reverse  the  sense. 

3.  When  both  members  of  an  inequality  are  positive,  we  may  raise 
to  any  power ; or  ive  may  extract  any  root,  provided  we  use  only  the 
positive  roots. 

For  example,  raise  both  members  of  7>5  to  the  second  power. 
We  have, 

19  >25. 

Extracting  the  square  root  of  this  last  inequality,  we  have, 

7>5; 

hut  we  could  not  take  the  negative  roots  and  say 

— 7>  —5. 

These  principles  will  enable  us  to  transform  an  inequality  so  as 
to  make  any  quantity  stand  alone  in  one  member,  greater  or  less  than 
a resulting  quantity  ; thus,  let  us  have, 

3z— 9>21. 


108 


ELEMENTS  OF  ALGEBRA. 


Transposing  and  dividing,  we  have, 


£>10. 


Examples. 


1.  2£  + 5>12. 

2.  ^-l<£  + 4. 


Ans.  £<10. 


Ans.  £>f. 


3.  f—  |<l0-6£. 
o 2 


Ans.  £<-rx  • 
o7 


100 


. U-C/ U/U 

Ans.  £> — = — — . 
d+1 


ac—cib 


a 


a 


$8.  Problems. 

A.  problem  is  a question  proposed  for  solution. 

The  solution  consists, 

1.  In  translating  the  given  conditions  into  algebraic  language, 
and  thus  deriving  one  or  more  equations,  involving  the  given  and 
the  required  elements  ; and 

2.  In  the  solution  of  the  resulting  equation  or  equations. 

Represent  the  required  elements  by  x,  y,  etc.,  and  then  use  these 

quantities  as  though  their  values  were  known. 

Be  careful  in  forming  an  equation  to  put  like  quantities  equal  to 
each  other;  that  is,  men  = men,  pounds=pounds,  etc.  Be  careful, 
also,  that  the  quantities  are  expressed  in  a common  unit. 

What  two  numbers  are  those  whose  sum  is  a,  and  whose 
difference  is  b ? 

Here,  although  two  numbers  are  required,  it  is  not  necessary  to  use 
more  than  one  unknown  quantity ; thus, 


Let  x—  the  greater. 
Then  a—x=  the  less. 


From  the  condition,  we  shall  have 


x—(a—x)  = b. 
2x—a  + b. 


X— — - — , the  greater ; 

A 


a + b 


EQUATIONS. 


109 


We  may  also  use  two  unknown  quantities  in  tlie  solution  of  this 
problem ; thus, 

Let  x = the  greater,  and 

y — the  less. 

Then  x+y=a, 

and  x—y—b', 

, a + b a b 

whence  x=— 

/v  A A 

a—b  a b 

y~ir~2~2' 

These  results  may  be  translated  into  English  thus : 

The  greater  of  two  quantities  is  equal  to  the  half  sum,  phis  the 
half  difference. 

The  lesser  of  two  quantities  is  equal  to  the  half  sum,  minus  the 
half  difference. 

2.  What  number  is  that  froniAvhich  if  5 be  subtracted,  two-thirds 
of  the  remainder  will  be  40  ? 

Let  cr=the  number. 

Then,  x— 5=the  remainder. 

From  the  conditions,  f(a— 5)  =40. 

x — 65. 

3.  A horse  said  to  a mule:  If  I give  you  one  of  my  sacks  we 
shall  have  an  equal  number;  if  I take  one  of  yours,  I shall  have 
double  the  number  you  have  left.  How  many  had  each? 

Let  x—  the  number  the  horse  had. 
y—  the  number  the  mule  had. 

Then,  x—l=y  + l, 

and  x + l = 2(y  — l), 

whence  x=7,  y— 5. 

4.  Divide  64  into  two  parts  which  shall  be  to  each  other  as  3 to  5. 

Let  3x=  one  part, 

Then  ox—  the  other. 

But  3.,r  + 5x=64. 

Whence,  x=8. 

3x=24  one  part, 

5.r=40  the  other. 


no 


ELEMENTS  OF  ALGEBRA. 


5.  Divide  a into  two  parts  which  shall  be  to  each  other  as  b to  c. 


Ans. 


ab  ac 
b + c’  l + c' 


6.  A man  left  £2,400  to  be  divided  between  two  sons  and  a servant. 

The  sons’  parts  were  to  be  to  each  other  as  3 to  2,  and  the  servant 
was  to  have  half  as  much  as  the  son  who  received  the  smaller  sum. 
How  much  had  each  ? Ans.  $1,200,  $800,  $400. 

7.  A person  upon  being  asked  his  age,  replied:  that  J of  his  age 

multiplied  by  -fa  of  his  age  would  give  a product  equal  to  his  age. 
IIow  old  was  he  ? Ans.  16  years. 

8.  A and  B have  the  same  income;  A contracts  an  annual  debt  of 

% of  his  income ; B lives  upon  £ of  his ; at  the  end  of  10  years,  B 
lends  A money  enough  to  pay  off  his  debts,  and  has  1607  left.  What 
was  their  income  ? Ans.  2S0I. 


9.  A man,  fifteen  years  after  his  marriage,  was  asked  the  age  of 
himself  and  of  his  wife  at  their  marriage.  He  replied  that  he  was 
then  twice  as  old  as  his  wife,  but  that  now  he  was  only  once  and  a 
half  as  old.  What  Avere  their  ages  ? Ans.  30  and  15. 


10.  A person  passed  £ of  his  age  in  childhood,  Th  in  youth,  i and 
5 years  besides  in  matrimony,  at  the  end  of  which  time  he  had  a son, 
Avho  died  4 years  before  his  father,  having  reached  half  his  father's 
age.  What  Avas  the  father’s  age  ? Ans.  S4. 


11.  A privateer  running  at  the  rate  of  10  miles  an  hour  discovers 
a ship-of-Avar  18  miles  off  pursuing  at  the  rate  of  8 miles  an  hour. 
IIow  many  miles  Avill  the  privateer  make  before  the  ship  overhauls 
her  ? 

Let  x—  the  distance  the  privateer  will  make  before  she  is  overhauled. 

Then,  ®+18=  the  ship’s  corresponding  distance. 

X 

— will  be  the  number  of  hours  required  by  the  privateer  to  reach  the  point  of 


union.  will  be  the  time  required  by  the  ship. 


Since  these  times  must  be  equal,  we  have, 
x ® + 18 
10 


8 

x=— 90. 


, whence. 


It  will  be  observed  that  our  result  is  negative.  IV e must,  then,  reckon  back- 
ward 90  miles  to  find  the  place  at  which  the  vessels  would  have  teen  together. 


EQUATIONS. 


Ill 


This  results  from  the-  fact  that  positive  and  negative  quantities  do  not  differ 
essentially,  but  merely  in  their  positions  with  respect  to  the  origin  of  reckoning. 
It  is  thus  that  the  algebra  takes  no  note  of  the  word  “ pursuing,”  except  in  its 
essential  sense  of  traveling  on  the  same  line  with  the  object  with  which  this 
relation  obtains.  So  in  general,  the  algebra  always  interprets  words  which  have 
relative  meanings  in  their  broadest  sense.  It  thus  makes  no  distinction  between 
“will  be”  and  “has  been,”  “up”  and  “ down,”  “ to  the  right  ” and  “ to  the 
left,”  etc. 

A negative  result  shows  that  there  has  been  some  error  made  in  the  use  of 
words  in  the  enunciation.  It  is  to  be  interpreted  in  a directly  contrary  sense 
from  a positive  result. 

12.  Find  two  numbers  such  that  tlie  first  added  to  three  times  the 

second  will  give  11 ; and  three  times  the  first  less  twice  the  second 
will  give  zero.  Ans.  2 and  3. 

13.  Find  three  numbers  such  that  three  times  the  first  increased 

by  twice  the  second  will  give  16  ; and  twice  the  third  less  the  first 
will  give  6 ; and  five  times  the  third  divided  by  twice  the  second  will 
give  10.  Ans.  2,  5,  and  4. 

14.  There  are  three  numerals  expressed  by  single  figures  'which 

added  together  give  8;  written  in  a certain  order  they  give  five  times 
the  number  expressed  by  the  second  and  third  written  together  ; if 
the  first  and  third  be  written  together,  they  express  f-  of  the  number 
formed  by  writing  the  first  and  second  together.  What  are  the  num- 
bers ? Ans.  1,  2 and  5. 

15.  What  three  numbers  are  those,  which  if  three  times  the  first 
be  added  to  twice  the  second  and  this  sum  divided  by  the  third,  will 
give  4;  and  if  three  times  the  second  be  added  to  five  times  the  third 
and  the  sum  be  divided  by  the  second,  will  give  7 ; and  if  four  times 
the  sum  of  all  the  numbers  be  divided  by  2,  will  give  11  ? 

Ans.  2,  5,  and  4. 

16.  A man  upon  being  asked  his  own  and  his  wife’s  ages,  replied 
that  his  age  divided  by  his  wife’s  would  give  one-fifteenth  of  her  age; 
but  that  if  30  had  been  first  subtracted  from  his  own  age,  the  result 
would  have  been  one-tliirtietli  of  his  wife’s  age.  How  old  was  each  ? 

Ans.  40  and  30. 

17.  The  length  of  a rectangular  field  exceeds  its  breadth  by  10 

chains,  and  it  has  2000  square  chains  in  it.  What  is  its  length  and 
breadth  ? Ans.  40  and  50. 

18.  What  two  numbers  are  those  which  if  the  first  be  taken  from 


112 


ELEMENTS  OF  ALGEBRA. 


the  second  and  the  difference  multiplied  by  the  first,  the  result  Mill 
he  four  times  the  square  of  the  first ; and  if  this  difference  be  squared, 
the  result  will  be  64  ? Ans.  2 and  10. 

19.  A merchant  has  two  sorts  of  tea,  one  worth  75  cents  a pound 

and  the  other  one  dollar.  He  wants  to  make  a mixture  of  25  pounds 
which  he  can  sell  at  90  cents.  How  many  pounds  of  each  must  he 
take?  Ans.  10  of  the  1st  and  15  of  the  2d. 

20.  A hare  is  50  leaps  before  a greyhound,  and  makes  4 leaps  to 
the  greyhound’s  three;  but  two  of  the  hound’s  leaps  are  equal  to 
three  of  the  hare’s.  How  many  leaps  must  the  greyhound  make  to 
catch  the  hare  ? 

In  sucli  a case  as  this,  we  must  first  fix  upon  a unit  of  distance,  which  here 
may  be  either  one  leap  of  the  dog  or  one  of  the  hare.  Let  us  take  the  length 
of  the  hare’s  leap  as  the  unit  ; then,  the  length  of  the  dog’s  leap  will  be  £ times 
that  of  the  hare’s. 

Now,  if  x be  the  number  of  leaps  the  hound  must  make,  the  hare  will  in  the 
same  time  make  leaps.  This  number  of  leaps  made  by  the  hare,  added  to 
the  number  of  leaps  it  is  in  advance,  will  be  the  distance,  in  our  assumed  unit, 
from  where  the  dog  is  to  the  place  where  the  hare  is  caught.  But  the  dog  will 
pass  over  a distance  equal  to  \x  ; that  is,  the  number  of  leaps  multiplied  by  the 
length  of  one  of  them  ; hence, 

ix+  50=%x. 

a:=300.  Ans. 

21.  Two  trains  are  traveling  towards  each  other  at  the  rate  of  20 

and  33  miles  per  hour,  respectively ; they  are  265  miles  apart.  How 
long  before  they  will  meet  ? Ans.  5 hours. 

22.  If  the  trains  had  been  traveling  at  the  rate  of  a and  b miles 

per  hour  and  had  been  c miles  apart,  how  long  would  they  take  to 
come  together  ? . c 

a + b 

From  this  formula,  tell  at  once  what  the  time  would  have  been 
had  the  rates  been  2 and  3,  and  the  distance  10.  Also,  if  the  rates 
had  been  \ and  ^ , the  distance  Also,  if  the  rates  had  been  5 and 
— 7,  the  distance  6. 

23.  Two  travelers  set  out  at  the  same  time  to  meet  each  other, 
being  154  miles  apart.  If  one  travels  at  the  rate  of  3 miles  in  2 
hours,  and  the  other  at  the  rate  of  5 miles  in  4 hours,  how  far  shall 
each  travel  before  they  meet,  and  after  what  time  ? 

Ans.  Distances,  84  and  70.  Time,  56  hours. 


SYMBOLS  0 AND  00  . — DISCUSSIONS.  113 

Let  letters  be  used  for  tlie  numerals,  and  the  results  he  applied  to  particular 
cases,  as  above.  So  with  any  of  the  preceding  or  following  problems. 

24.  A can  do  a piece  of  work  in  5 days,  .and  B can  do  the  same 
in  7;  working  together,  how  long  will  they  be  at  it  ? 

Ans.  2ih  days. 

25.  The  half  of  A’s  fortune  less  B's  is  equal  to  83,000;  B’s  for- 
tune less  one-fifth  A’s  gives  zero.  What  sum  had  each  ? 

Ans.  A,  810,000,  B,  82,000. 

26.  What  two  numbers  are  those  which  cubed  and  added  together 

give  35,  and  which  if  the  first  be  squared  and  multiplied  by  the  sec- 
ond, and  this  result  be  added  to  the  product  of  the  second  squared  by 
the  first,  will  give  30  ? Ans.  2 and  3. 


SECTION  VIII. 

SYMBOLS  O AND  oo.— DISCUSSIONS. 

99.  The  entire  absence  of  value  is  represented  by  the  symbol  0, 
called  zero.  A man  who  is  altogether  destitute  of  money,  has  0 
dollars.  This  is  the  ordinary  meaning  of  zero ; but  in  algebra  it 
has  a somewhat  more  extended  signification. 

A man  may  not  only  be  destitute  of  money,  but  he  may  have  less 
than  none  at  all,  in  an  algebraic  sense,  as  we  have  seen  in  Art.  26; 
he  may  be  in  debt.  In  such  a case,  zero  is  the  point  of  division  be- 
tween his  assets  and  liabilities. 

Again,  if  we  agree  to  reckon  distances  upon  a right  line  or  scale 
from  a particular  point  upon  it,  such  point,  having  no  distance  from 
the  origin  of  distances,  is  the  zero  point.  We  have  a familiar  ex- 
ample of  this  in  the  thermometer ; the  position  of  the  zero  point 
upon  the  scale  being  a mere  matter  of  agreement. 

Thus,  zero  in  algebra  is  but  the  point  of  separation  between  all 
positive  quantities  on  the  one  hand,  and  all  negative  quantities  on 
the  other.  If  a positive  quantity  be  continually  diminished,  it  will 
approach  zero  as  a limit ; but  as  we  cannot  conceive  of  a quantity  so 
small  that  there  cannot  be  a smaller,  we  could  never  find  the  smallest 
possible  quantity.  But  that  which  continually  approaches  a limit 
may  be  said  to  have  the  limit  itself  for ’its  extreme  value;  and  so 
n 


114 


ELEMENTS  OF  ALGEBRA. 


0 is  often  said  to  be  a quantity  smaller  than  any  assignable  quan- 
tity. 

The  symbol  co , called  infinity,  represents  a quantity  than  which 
a greater  is  impossible.  It  is  the  limit  of  all  augmentation.  We 
may  approach  it  to  any  possible  degree  of  approximation,  but  cau 
never  reach  it. 

100.  Combinations  of  0 and®. 

Any  product  which  has  0 in  it  as  a factor,  must  itself  be  zero  ; thus, 

a x 0 = 0. 

For  if  zero  be  taken  any  number  of  times,  it  will  still  be  zero  ; or 
if  any  quantity  be  taken  zero  times,  we  shall  have  zero  as  well. 

Zero  divided  by  any  finite  quantity  is  equal  to  zero ; thus, 


a 


For,  since  the  product  of  the  divisor  and  quotient  must  produce 
the  dividend,  we  must  have  0 for  the  quotient  to  give  with  a,  the 
dividend  0 ; or,  more  briefly,  zero  times  the  fractional  unit,  gives 
zero. 

A finite  quantity  divided  by  zero,  gives  infinity ; thus, 

a 

--  co. 

For,  as  we  diminish  the  divisor,  we  increase  the  quotient.  When 
we  have  made  the  divisor  the  least  possible,  the  quotient  must  be 
the  greatest  possible,  that  is,  infinite. 

Zero  divided  by  zero,  gives  any  quantity  whatever,  or,  as  it  is 
said,  is  Indeterminate  j thus, 

0 

Q-a. 

For,  any  quantity,  a,  the  quotient,  multiplied  by  the  divisor  0, 
gives  0 the  dividend. 

Any  quantity  divided  by  infinity  gives  zero  ; thus, 

— = 0. 
oo 

For,  as  the  divisor  is  increased,  the  quotient  is  diminished. 
When  the  divisor  is  the  greatest  possible,  the  quotient  must  be  the 
least  possible,  or  zero. 

Infinity  divided  by  a finite  quantity  gives  infinity ; thus, 

oo 

• = 00  . 

a 


SYMBOLS  0 AND  CO  .—DISCUSSIONS. 


115 


For,  when  the  dividend  is  the  greatest  possible,  the  divisor  being 
constant,  the  quotient  must  also  be  the  greatest  possible,  or  infinite. 
Thus  we  may  say  that, 

1.  Zero  multiplied  by  any  quantity  is  zero; 

2.  Zero  divided  by  a finite  quantity  is  zero; 

3.  A finite  quantity  divided  by  zero  is  infinite; 

4.  Zero  divided  by  zero  is  any  quantity ; 

5.  A' finite  quantity  divided  by  infinity  is  zero; 

6.  Infinity  divided  by  a finite  quantity  is  infinite. 


101.  "Vanishing:  Fractions. 


1.  Sometimes  an  algebraic  fraction  reduces  to  the  form  under  a 
certain  hypothesis  made  upon  the  quantities  which  enter  it,  when 
0 is  not  the  true  value  of  the  fraction  ; thus, 

(a3—x3)2 


when  a — b,  becomes, 


This  is  not  the  true  value  of  y ; for,  before  making  the  hypothesis, 
resolve  the  numerator  and  denominator  into  their  factors  ; thus, 
_(a—x)(a2  + ax  + x2) 

^ {a  + x)  (a — x)  ’ 


and  cancel  the  common  factor,  x—a\  thus, 

a 2 + ax+x2 

y— 

J a + x 

Now  make  a=b,  and  we  have, 

3 a2  3 a , , ,, 

y~~. — =— , the  true  value  ot  y. 
'Za  Z 

2.  Again, 

_ f a2  —x2  \ __  0 

y~\{a-x)*)IB=a*~  0* 


But, 

3.  Again, 


_(a  + x)(a— x)  _(a  + x\  _2  a 
y—{a—x){a—x)~~\a—x)x=a,  0 


* This  notation  shows  that  a is  to  be  made  equal  to  b in  the  expression  ; read,  when  x is 
equal  to  a. 


116 


ELEMENTS  OE  ALGEBRA. 


But, 


y- 


(a—x)(a—x) 


( a—x_ \ _ 0 _ 

\az +ax  + x2')  x=a~  3a2~ 


(a—x)(a2+ax  + x2)  Xa^+ax  + x^j 

Such  expressions  are  called  Vanishing  Fractions.  They  appear 
to  be  equal  to  for  a particular  hypothesis,  but  really  are  not  so. 
They  reduce  to  this  form  from  the  presence  of  a concealed  common 
factor  which  becomes  zero  under  the  particular  hypothesis.  When 
such  factor  is  stricken  out,  the  true  value  results  under  the  hypoth- 
esis. As  we  have  seen,  the  true  value  may  be  either  7,-,or 

J b’  O’  a 


102.  The  Problem  of  the  Couriers. 

The  discussion  of  an  expression  consists  in  making  every  possible 
supposition  upon  the  arbitrary  quantities  which  enter  it,  and  inter- 
preting the  results. 

The  following  problem  gives  rise  to  some  interesting  results. 

Two  couriers  traveling  on  the  same  straight  line,  one  at  the  rate 
of  m miles  an  hour,  and  the  other  at  n miles  an  hour,  are  separated 
by  a distance  of  a miles  at  12  o’clock  m.  When  will  they  be 
together  ? 

A B 


m n 

Let  one  of  the  couriers  be  at  A and  the  other  at  B.  The  distance 
from  A to  B will  be  a. 

Now,  let  the  couriers  be  moving  to  the  right,  and  let  us  reckon 
all  distances  from  A as  the  origin  of  distances, — those  to  the  right 
of  A being  positive,  and  those  to  the  left  negative. 

Let  the  courier  at  A be  traveling  at  m miles  an  hour,  and  the  one 
at  B at  the  rate  of  n miles  an  hour.  Using  two  unknown  quanti- 
ties, let  t be  the  number  of  hours,  counting  from  12  M.  until  the 
union  takes  place,  and  x the  number  of  miles  to  be  traveled  by  the 
courier  at  B to  reach  the  place  of  meeting.  Then  we  shall  have, 

nt—x  - - - (1), 

and  mt—xA-a  - - (2). 

Combining  these  equations  and  eliminating  x,  we  have, 

mt—nt—a. 

a 

W hence,  t — . 

’ m—n 


SYMBOLS  0 AMD  GO  . — DISCUSSIONS. 


117 


This  value  of  t in  (1)  gives, 

na 

X~  m—n 


for  the  distance  traveled  by  the  foremost  courier.  Now,  let  us  take 
the  root,  t=——^f  aud  make  every  possible  hypothesis  upon  a,  m, 
and  n,  and  ascertain  what  t will  denote  in  each  case. 

I.  First  let  &l>e  a positive  quantity,  and  m>n. 

In  this  case  the  denominator  of  the  fraction  which  expresses  the 
value  of  t,  will  be  positive,  and  since  a is  also  positive,  t must  be 
essentially  positive. 

We  shall  have  to  add  t hours,  therefore,  to  12  M.  in  order  to  find 
the  time  at  which  the  couriers  will  be  together. 

This  is  what  common  sense  would  tell  us,  since  we  simply  have 
the  case  of  one  courier  pursuing  the  other  at  a more  rapid  pace,  and 
so  must  be  constantly  gaining  on  him  and  at  last  must  overtake 
him. 


II.  Let  a he  positive,  and  m<n. 

In  this  case  the  denominator  of  the  fraction  is  negative  and  the 
numerator  positive ; t is,  therefore,  negative. 

We  must,  hence,  subtract  t hours  from  the  origin  of  time,  12 
o’clock;  that  is,  reckon  backwards  to  find  the  time  at  which  the 
couriers  were  together .* 

This  we  can  readily  understand,  also;  for  since  the  courier  in  ad- 
vance is,  under  the  hypothesis,  traveling  more  rapidly  than  the  one 
in  the  rear,  a moment  before  12  o’clock,  there  was  less  distance  be- 
tween them,  and  less  the  moment  before  that,  and  so  there  must 
have  been  a time  when  they  were  together. 

III.  Let  a he  positive,  and  m=n. 

In  this  case  the  denominator  of  the  fraction  will  be  0 : hence 
i is  qo  . 

This  is  plainly  as  it  should  be,  since  with  a certain  distance  be- 
tween them,  and  traveling  at  the  same  rate,  they  cannot  be  together 
in  any  finite  time.  The  result  go  is  here  equivalent  to  never. 

IV.  Let  a he  negative,  and  m>n. 

In  this  case,  the  denominator  of  the  fraction  is  positive,  but  the 
numerator  negative,  t is,  therefore,  negative.  We  must  reckon 
backwards  to  find  the  time  of  conjunction. 


* See  remarks  uuder  problem.  11,  page  110. 


118 


ELEMENTS  OF  ALGEBRA. 


a being  negative,  places  the  courier  who  is  traveling  at  the  rate 
of  n miles  an  hour,  m rear  of  the  other:  and  since  he  is  moving 
more  slowly,  they  must  have  been  already  together  in  the  past. 

V.  Let  a be  negative,  and  nr<n. 

t will  be  positive,  and  the  meeting  has  yet  to  take  place. 

YI.  Let  a be  negative,  and  m=n. 

t is  infinite,  and  may  be  either  positive  or  negative.  The  couriers 
never  have  been,  and  never  will  be  together;  or  we  may  say  they 
were  together,  and  that  they  will  be  together  an  infinite  time  either 
in  the  past  or  future. 

VII.  Let  a=0,  and  m>n  or  <n. 

t is  zero;  and  the  couriers  were  together  at  12  o’clock.  They 
manifestly  never  could  have  been  together  before,  and  can  never  be 
again. 

VIII.  Let  a=0,  and  m — n. 

t becomes 

That  is,  any  time  may  be  added  to  or  subtracted  from  the  epoch, 
12  o’clock. 

Since  the  couriers  are  together  at  12  o’clock,  and  are  moving  at 
the  same  rate,  they  have  always  been,  and  will  always  be  together. 

IX.  Let  a=0,  m=0,  and  n=0. 

Under  this  hypothesis,  we  have  again  t=%.  Here  the  couriers  are 
together  and  are  not  moving  at  all.  Of  course  they  always  have 
been  and  always  will  be  together. 

X.  "We  may  now  fix  the  value  of  any  three  of  the  four  quantities, 
t,  a,  in,  or  n,  at  pleasure,  and  thus  determine  the  remaining  one. 
Let  the  instructor  make  such  hypotheses  and  require  the  student  to 
determine  and  interpret  the  results. 

We  have  not  yet  entirely  exhausted  the  possible  hypotheses  upon 
the  quantities  which  enter  the  expression  under  consideration.  Ye 
may  make  in  and  n negative  in  succession;  in  which  case  the  couri- 
ers would  be  traveling  in  opposite  directions;  or  we  may  make  them 
both  negative  at  the  same  time,  in  which  case  they  would  be  travel- 
ing to  the  left.  Let  the  instructor  give  such  exercises. 

O 

103.  The  Problem  of  the  Lights. 

The  discussion  of  the  Problem  of  the  Lights,  as  it  is  called,  also 
gives  rise  to  many  interesting  and  instructive  results. 

The  problem  is  this : 


SYMBOLS  0 AND  00  . — DISCUSSIONS. 


119 


Given,  two  lights,  placed  anywhere  upon  a straight  line,  to  deter- 
mine the  point  or  points  which  will  be  equally  illuminated  by  each. 

We  must  first  know,  obviously,  the  law  which  governs  the  illumi- 
nating power  of  light;  and,  upon  this  point,  physics  teaches  us, 
that. 

The  intensity  of  n light  at  any  distance  is  equal  to  its  intensity  at 
the  distance  1,  divided  by  the  square  of  the  given  distance  j that  is, 
if  a light  has  an  intensity,  say  10,  at  the  distance  of  one  yard  from  it, 

at  two  yards  its  illuminating  power  will  be  ^ ; at  three  yards,  ^ ; 


at  x yards, 


10 


Then  let  the  lights  in  question  be  placed  upon  the  line  A B,  one 
at  A,  and  the  other  at  B. 

A c P B 


Let  the  distance  between  them  be  c,  and  let  the  intensities  of  the 
lights  at  the  unit  of  distance  be,  respectively,  a and  l.  Suppose  the 
point  P to  be  one  point  equally  illuminated,  and  let  its  distance  from 
A,  the  origin  of  distances,  be  x\  its  distance  from  B will  be  c—x. 
Let  distances  to  the  right  of  A be  positive ; those  to  the  left  will  be 
negative. 

Now,  the  illuminating  power  of  the  light  A,  for  the  point  P,  will 


be  ~ ; and  that  of  the  light  B for  the  same  point,  will  be^— — 2. 


Since  the  quantity  of  light  from  each  of  the  lights  for  this  point  is 
to  be  the  same,  we  must  have 

a _ b 
x2  (e — x)2 

Solving  this  equation,  we  have 

cV  a , 

x— — — —>  and 

Va  + yb 

cVa 

gr “ 

Va  — Vb 

Since  we  find  two  roots  of  the  equation,  there  must  be  two  points 
of  equal  illumination,  so  long,  at  least,  as  the  roots  are  of  different 
values. 


120 


ELEMENTS  OF  ALGEBRA. 


Now,  to  discuss  these  two  expressions, 

I.  Let  c be  positive,  and  a >b. 


The  first  root  will  manifestly  be  positive ; the  point  corresponding 
to  this  value  of  x will,  therefore,  be  found  somewhere  to  the  right  of 

a/ Cl 

A.  Its  distance  from  A will  be  c times  the  fraction  - — — — — But 

Vn  + V# 

since  a is  greater  than  b in  this  case,  the  denominator  is  less  than 
twice  the  numerator,  and,  hence,  this  fraction  is  greater  than  c 
times  the  fraction  must,  therefore,  be  greater  than  \c.  x is  thus 

Q 

greater  than  -,  or  the  point  P must  be  farther  from  A than  it  is  from 

/y 


B. 

This  is  what  common  sense  would  teach;  for,  the  stronger  light 
being  at  A,  the  point  equally  illuminated  must  be  nearer  the  lesser 
light. 

The  second  root  under  this  hypothesis  is  also  positive,  and  since 

in  the  fraction, — - — — , the  denominator  is  less  than  the  numerator, 

's/a—  Vb 

the  fraction  will  be  greater  than  unity,  and  consequently  c times  it 
will  be  greater  than  c.  x is  thus  greater  than  c ; that  is,  this  second 
point  lies  beyond  B,  the  feebler  light. 

This  must  be  so,  since,  as  we  move  to  the  right  of  the  lesser  light, 
the  difference  between  the  quantity  of  light  from  the  two  sources 
becomes  less  and  less.  It  will,  thus,  at  last  be  zero  ; that  is,  we 
shall  reach  a point  which  will  receive  the  same  quantity  of  light 
from  each. 


II.  Let  c be  positive  and  a<b. 

Under  these  hypotheses,  the  first  root  will  still  be  positive,  but  it 
will  be  less  in  value  than  c;  that  is,  the  corresponding  point  will  lie 
nearer  to  A,  now  the  feebler  light. 

The  second  value  of  x will  be  negative ; that  is,  the  second  point 
will  lie  to  the  left  of  A. 

In  this  case  we  have  but  changed  the  places  of  the  lights,  and  of 
course  the  circumstances  are  just  reversed  from  those  in  the  pre- 
vious case. 


III.  Let  c be  positive  and  a— b. 

In  this  case  the  first  root  becomes,  x— — 5 that  is,  the  first 

2 Va  2 


SYMBOLS  0 AND  CO  . — DISCUSSIONS.  121 

point  is  midway  between  the  lights.  Since  the  lights  are  equal, 
under  the  hypothesis,  this  is  manifestly  true. 

The  second  root  becomes, 

c/\/~a 

X~~0  = °°* 

The  farther  we  recede  from  the  lights,  the  less  will  be  the  differ- 
ence in  the  quantity  of  light  at  any  point ; but  this  difference  will 
not  become  zero,  or,  in  other  words,  the  quantity  of  light  will  not  be 
entirely  equal  until  we  have  reached  an  infinite  distance,  rvhich,  of 
course,  can  never  be. 

IV.  Let  c he  negative  and  a > o?’  < b,  or  a= b. 

The  effect  of  making  c negative,  is  to  place  the  light  whose  inten- 
sity is  b on  the  left  of  the  other,  the  light  having  the  intensity  a 
being  still  the  origin  of  distances.  The  separate  discussion  of  the 
several  relations  of  a and  h to  each  other  in  this  Case  would  give  the 
reverse  of  the  results  already  considered. 

V.  Let  c = 0 and  a>  or  <b. 

Both  roots  in  this  case  become  0,  and  thus  the  origin  is  equally 
illuminated. 

[The  supposition  that  c is  0,  places  the  two  lights  together  at  the 
origin.  The  lights,  however,  are  of  different  powers.  It  is  thus  laid 
down  in  several  late  works  on  the  subject,  that  this  point  cannot  be 
equally  illuminated  by  the  two  lights,  and  consequently  that  the 
analysis  fails  for  this  case.  Several  attempts  have  been  made  to  ex- 
plain the  anomaly ; but  in  reality  it  should  seem  that  , there  is  no 
failure  and  no  real  difficulty. 

Let  it  be  remembered  that  our  equation  is  deduced  under  a postu- 
late borrowed  from  physics  with  regard  to  the  effect  of  light  at  dif- 
ferent distances.  The  algebra  accepts  that  law  as  absolutely  true, 
and  its  results  must  be  interpreted  strictly  under  that  hypothesis. 

Let  us  then  approach,  say,  the  light  whose  intensity  is  a at  the  unit’s 
distance.  When  we  have  reached  the  distance  -V,  its  intensity  is,  by 

the  law,  at  the  distance  T-^g-,  it  is  10,000«;  at  0 it  is  go. 

The  light  whose  intensity  is  5,  or  has  any  other  intensity  at  the  unit’s 
distance,  is  also  oo  at  the  point  zero.  Since,  then,  under  our  assumed 
law,  both  lights  are  infinite  at  the  point  0,  that  point  is  equally  il- 
luminated, and  the  roots  are  truly  zero.  The  results  of  the  analysis 


122 


ELEMENTS  OF  ALGEBRA. 


may  be  farther  vindicated  by  using  mathematical  lines,  and  thus  re- 
moving the  question  entirely  beyond  the  laws  of  physical  science. 
See  note  at  the  end  of  the  book.] 

VI.  Let  c— 0 and  a=b. 

The  first  root  becomes  0,  and  the  second  -gq  that  is,  One  of  the 
points  is  at  the  origin,  and  the  other  anywhere  we  please. 

This  is  easily  understood. 

VII.  Let  a and  b be  negative  in  succession,  or  together. 

In  this  case  both  roots  are  imaginary. 

This  is  as  it  ought  to  be,  since  under  our  assumed  law  of  physics 
the  absence  of  light,  or  total  darkness,  cannot  be  reached  until  the 
distance  from  the  source  becomes  infinite.  There  is  thus  an  infinite 
separation  between  absolute  light  on  the  one  hand,  and  no  light  at 
all  on  the  other.  Less  than  no  light  is,  therefore,  impossible. 

VIII.  Let  a=0,  b = 0,  c=0. 

Both  roots  become  -g,  and  thus  all  points  are  equally  illuminated, 
and  this  must  be  true  since  no  point  would  have  any  light  at  all. 

104.  General  Properties  of  Equations  of  the  Second  Degree. 

Equations  of  the  second  degree  containing  but  one  unknown 
quantity  possess  some  important  properties  which  we  shall  now  pro- 
ceed to  investigate. 

Besummg  the  general  equation, 

Xs +2  px=q (1), 

let  us  complete  the  square  and  transpose  all  the  terms  into  the  first 
member.  We  shall  have, 

xz  + 2g>x+p~  — (q+p-)—0. 
or,  (x+jp)2  — (q+p2)= 0. 

Regarding  this  as  the  difference  of  two  squares,  we  may  write, 
(x+2J  — Vq+]J2){x+p+Vq+P’)  = 0 (2). 

This  equation  can  be  true  only  upon  the  supposition  that  one  of 
the  factors  composing  the  first  member  is  equal  to  zero. 

We  may,  then,  have  either, 

x+p—\/q+p2= 0,  or, 

x+jj  + Vq+2)2  = 0. 


SYMBOLS  0 AND  CO  . — DISCUSSIONS.  123 

From  these  we  get, 

x—  —p  + VqTp* : the  first  root,  and 

x^—p—'f q+p 2 : the  second  root. 

As  there  are  no  other  possible  values  of  x which  will  satisfy  this 
equation,  we  may  say  that, 

1st.  Every  equation  of  the  second  degree  has  tioo  roots  and  only  two. 
By  inspection  of  equation  (2)  we  see  that, 

2d.  The  first  member  of  every  equat  ion  of  the  second  degree,  when 
the  second  member  is  0,  is  composed  of  two  factors,  having  the  un- 
Tcnown  quantity  for  the  first  term  in  each,  and  the  respective  roots 
with  their  signs  changed  for  the  other  terms. 

If  the  roots  are  given,  the  equation  can  be  constructed  at  once ; 
thus,  for  example,  let  the  roots  of  a certain  equation  be  x —a  and 
x=—b.  The  equation  will  be, 

[x— a)(x  + b)  =0,  or 
x2  + bx—ax—ab—0. 

Examples. 

The  roots  being  4 and  — 5 ; what  is  the  equation  ? 

Ans.  x2  — x— 20  = 0. 

The  roots  being  ab  and  — c:  a and  a2:  — 7 and  8:  —6  and  —5: 

^ and—-.:  —land  +1:  V — 1 and  — V — 1:  what  are  the  equa- 
b d 

tions  in  these  several  cases  ? 

105.  The  Sum  of  the  Roots. 

Let  us  now  add  together  the  two  roots, 
x'  — —p  + V q +p 3 
x"=—p—Vq+p2 

using  x'  and  x!'  to  distinguish  them.  We  get, 
x'  -\-x"  ——2 p. 

Hence  we  may  say,  that, 

3d.  The  sum  of  the  two  roots  is  equal  to  the  co-efficient  of  the  first 
power  of  the  unlcnoivn  quantity  with  its  sign  changed. 


124 


ELEMENTS  OF  ALGEBEA. 


106.  The  Product  of  the  Hoots. 

Now,  multiplying  the  roots  together, 
x'  =—p  + */q+p* 

x"~  —p—  V q +p2 ; we  have, 
x'x" = —q. 

Hence,  we  may  say  that, 

Uh.  The  p>roduct  of  the  two  roots  is  equal  to  the  second  member 
with  its  sign  changed . 


107.  The  Greatest  Numerical  Value  of  q when  Negative. 

Since  the  product  of  the  two  roots  is  always  equal  to  q (the  second 
member),  with  a contrary  sign,  if  q is  negative,  the  roots  must  have 
like  signs;  and  when  added  together,  the  algebraic  sum  will  be  their 
numerical  sum.  Now,  this  sum  is,  as  we  have  seen,  equal  to  2p. 

We  thus  have  the.  sum  of  two  quantities  given:  and  now  let  us 
find  how  to  divide  this  sum  into  two  parts,  so  that  their  product 
shall  be  the  greatest  possible. 

Let  d be  the  difference  between  the  two  parts,  2p  being  the  sum. 

Then,  the  greater  will  be  (Prob.  1,  Art.  98), 


and  the  less, 

d 

2’ 

Their  product  must  be  equal  to  q,  and  we  shall  have, 


Now,  as  d is  diminished,  q will  increase,  until  when  <7=0,  q will  be 
equal  to  p2,  and  will  then  be  the  greatest  possible:  that  is  to  say, 
when  the  roots  are  equal,  their  product  will  be  the  greatest  possible. 
q,  therefore,  can  never  be  greater  than  pz.  This,  however,  is  under 
the  supposition  that  q is  negative. 

When  q is  positive,  since  the  roots  multiplied  together  must  then 
give  — q,  they  will  have  contrary  signs,  and  when  added,  will  give 
their  numerical  difference,  instead  of  their  numerical  sum.  There  is, 
then,  no  limit  to  the  values  of  q in  such  a case. 

Hence,  we  may  say,  that, 


SYMBOLS  0 AND  00  . — DISCUSSIONS. 


125 


5 th.  When  the  second  member  is  negative,  it  can  never  be  numeri- 
cally greater  than  the  square  of  one  half  the  co-efficient  of  the  first 
power  of  the  unknown  quantity. 


108.  Discussion  of  the  Four  Forms. 

Resuming  the  Four  Forms  already  given  (Art.  82),  viz. : 


x2 -\-2px~  q - - - - (1). 

x2—2px—  q - - - - (2). 

x2+2 px——q  - - - - (3). 
x2—2  px=—q  - - - - (1). 

Writing  the  root,  respectively,  we  have, 

x=—p±Vq  +p'2  - - - (1). 
x=+p±  Vq+p2  - - - (2). 
x——p±V—q+p3  - - (3). 
x—+p±V—q+p'i  - - (4). 


Since  V" q +p 2 is  greater  than  p,  the  radical  parts  of  the  roots  in 
the  first  and  second  forms  are  greater,  numerically,  than  their  entire 
parts.  The  radical  parts  will  therefore  govern  the  signs  in  these 
two  forms ; so  that  in  the  first  and  second  forms  the  signs  of  the 
roots  will  be  unlike,  and  the  negative  root  will  be  numerically  the 
greater  in  the  first,  while  the  positive  root  will  be  the  greater  in  the 
second. 

In  the  third  and  fourth  forms,  the  roots  will  be  imaginary  when 
q is  numerically  greater  than  p2.  This  should  be  the  case,  since  we 
have  proved  that  it  is  impossible  for  q,  when  negative,  to  be  greater 
thany>2.  When  q has  such  a value,  it  shows  that  the  equation  from 
which  the  roots  came  is  impossible. 

When  q is  less  than  p2,  the  radical  parts  in  the  last  two  forms  will 
be  less  than  p,  the  entire  parts  ; so  that  when  the  roots  are  real  in 
the  third  and  fourth  forms,  they  are  both  negative  in  the  third,  and 
both  positive  in  the  fourth. 

Ifg  — p%  in  the  third  and  fourth  forms,  the  radical  parts  of  the 
roots  reduce  to  zero.  The  roots  will  then  be  equal  in  either  form, 
both  being  —p  in  the  third,  and  +p  in  the  fourth  form. 

When_p=0,  the  roots  in  the  first  two  forms  reduce  to  ±Vq,  and 
in  the  last  two  ± V —q.  Under  this  supposition,  the  roots  are  thus 
equal  with  contrary  signs  in  the  first  and  Second  forms,  and  are 


126 


ELEMENTS  OF  ALGEBRA. 


always  imaginary  in  the  third  and  fourth.  This  ought  evidently  to 
be  the  case,  since  by  making  p=0  in  the  forms  themselves,  they  re- 
duce to  two  sets  of  incomplete  equations;  thus, 

x 2 — q 
x2  — — q. 

Making  q— 0,  the  roots  of  the  first  and  third  forms  become, 

x — —p  ±p. 

Those  of  the  second  and  fourth, 

x=+p±p. 

Or,  x—  0,  and  x—  —2 'p 

x=+2 p,  and  x=  0. 

Thus,  under  this  hypothesis,  one  of  the  roots  in  each  form  be- 
comes zero. 

The  reason  of  this  readily  appears,  for  if  q be  made  0 in  the  forms 
themselves,  we  have,  for  the  first  and  third, 

x2  + 2px=0 ; - - (a) 

and  for  the  second  and  fourth, 

x2—  2px=0.  - - (b) 

These  may  be  written, 

x(x  + 2p)=Q 
x(x—2p)=0. 

We  may  satisfy  these,  by  making  either  one  of  the  factors  zero  ; 
thus, 

x—0  or  (x  + 2p)  — 0 
x—0  or  (x—2p)=0. 

Whence  the  roots  are, 

,r=0  and  X——2 p 
x=0  and  x—  +2p. 

We  might  at  once  divide  out  x from  equations  (a)  and  (b),  and 
thus  reduce  them  to  equations  of  the  first  degree;  and,  generally, 
whenever  ivc  can  divide  one  equation  through  by  the  unknown  quan- 
tity, one  of  its  roots  is  zero. 

If  p = 0 and  <7=0,  the  roots  are  0. 

This  supposition  reduces  the  equations  themselves  to 

x2=0  .•.  x=0  and  x=0. 

Much  the  same  discussion  may  be  had  from  the  forms  themselves, 


ARITHMETICAL  PROGRESSION.  127 

by  the  use  of  the  3d  and  4th  general  properties  established  in  Arti- 
cles 105  and  106. 

For  example,  in  the  first  form  the  roots  must  hare  unlike  signs, 
because  their  product  must  give  — q;  they  are  unequal  in  numerical 
value,  and  the  negative  is  the  greater,  because  their  algebraic  sum 
must  give  —2 p. 

Let  the  student  carry  on  the  discussion. 


o»o- 

SECTION  IX. 

ARITHMETICAL  PROGRESSION.— RATIO  AND  PROPORTION.- 
GEOMETRICAL  PROGRESSION. 

109.  A series  is  a succession  of  terms,  any  one  of  which  may  be 
derived  from  the  preceding  term  or  terms  according  to  a uniform  law, 
called  the  law  of  the  series. 

110.  Arithmetical  Progression. 

An  Arithmetical  Progression  is  a series  in  which  any  term  may  be 
derived  from  the  one  preceding  it  by  adding  a constant  quantity, 
called  the  common  difference  ; thus, 

1,  3,  5,  7,  9 - - etc., 

is  such  a progression,  the  common  difference  being  2.  When  the 
common  difference  is  positive,  as  in  this  case,  we  have  an  increasing 
progression.  When  the  common  difference  is  negative,  we  shall  have 
a decreasing  progression ; thus, 

9,  7,  5,  3,  1,  -1,  -3  - - etc., 
is  a decreasing  progression,  in  which  —2  is  the  common  difference. 

111.  Formula  for  the  Last  Term. 

Any  term  with  which  we  choose  to  begin  the  series  is  called  the 
first  term;  that  with  which  we  end  it,  is’calledthe  last  term.  These 
two  are  the  extremes. 

Let 

a,  b,  c,  e,  f - - etc., 

be  an  arithmetical  progression,  in  which  cl  is  the  common  difference. 
Then  we  must  have, 

b=ci  + cl,  c=b  + d=a  + 2d, 
e—c  + cl—a  + od;  etc., 


128 


ELE SCENTS  OE  ALGEBRA. 


and  it  is  evident  that  any  term  may  be  found  by  adding  to  the  first 
term  as  many  times  the  common  difference  as  there  are  preceding 
terms. 

Then  if  l represent  the  last  term,,  and  n the  number  of  terms,  we 
must  have, 

l=a+(n— l)d. 

If  d be  negative,  the  formula  will  be, 

l—a—(n—l)d. 

Hence,  we  may  say  that, 

To  find  any  term  of  an  arithmetical  progression,  multiply  the  com- 
mon difference  by  the  number  of  preceding  terms,  and  add  this  prod- 
uct to  the  first  term,  if  the  progression  is  increasing,  or  subtract  it 
if  it  is  decreasing. 


112.  The  Sum  of  Equi-distant  Means. 

Let  us  have  an  increasing  progression  of  a definite  number  of 
terms.  If  t represent  the  term  which  has  m terms  before  it,  and  d 
the  common  difference,  we  shall  have, 

t—a  + md  - - - - (1). 

How,  if  we  reverse  the  order  of  terms,  we  shall  have  a decreasing 
progression,  with  —cl  for  the  common  difference.  If  /'  represent  the 
term  which  now  has  m terms  before  it,  we  shall  have 

t'—l—md  - - - - (2). 

Adding  (1)  and  (2),  we  have 

t -\-t  — a "h  l. 

Hence,  we  conclude,  that 

In  any  arithmetical  progression,  the  sum  of  the  two  terms  which 
are  at  equal  distances  from  the  extremes,  is  equal  to  the  sum  of  the 
extremes  themselves. 

113.  Formula  for  the  Sum. 

If  s represent  the  sum  of  n terms  of  a progression,  we  shall  have, 
s=-a  + b + c+  - - - - j+lc+l. 

Reversing  the  order,  we  shall  have, 
s=l-\-lc  + 


c-  + b + a. 


ARITHMETICAL  PROGRESSION. 


129 


Adding  these  equations,  member  to  member,  we  have, 

2s—  (a  + 7)  + (7+ 7c)  + ...  - (&  + 5)  + (7  + ft). 

The  terms  taken  two  and  two,  as  shown,  give  equal  sums,  from  the 
principle  just  established;  and  there  will  be  as  many  of  these  partial 
sums  as  there  are  terms  in  the  progression ; so  that  we  shall  have, 

2 s=  ( a + l)n;  whence 
(a  + l)n 

Hence,  we  may  say,  that, 

The  sum  of  a definite  number  of  the  terms  of  an  arithmetical  pro- 
gression, is  equal  to  half  the  sum  of  the  extremes,  multiplied  by  the 
number  of  terms. 

The  two  formulas, 

l=a  + {n—T)d  - - - - (1), 
s={a+T)n  ....  (2), 

are  sufficient  to  solve  all  ordinary  questions  touching  an  arithmetical 
progression.  There  are  altogether  five  arbitrary  quantities, 

a,  l,  d,  n,  s, 

entering  these  formulas.  We  may  assume  any  three  of  them  at  pleas- 
ure, and,  regarding  the  other  two  as  unknown,  we  may  combine  the 
equations,  and  thus  deduce  their  separate  values. 

For  example,  let  d,  n,  and  s be  given  to  find  a and  7. 

Substitute  the  value  of  l from  (1)  in  (2),  we  shall  have, 

_[a  + a + {n— 1 ) d~\  n 

s-  - . 

From  this, 

_ 2s—  (n— l)dn 
a~~  2 n 

This  value  of  a in  (1),  gives, 

7_  2s  + (n— 1)  dn 
~~  ~2n  ' 

Let  the  student  be  required  to  find  the  formulas  for  determining 
any  two  of  the  elements  when  the  other  three  are  given  by  the 
instructor. 

H 


130 


ELEMENTS  OF  ALGEBRA. 


Examples. 

1.  In  the  progression  1,  2,  3,  etc.,  of  14  terms,  what  is  the  last 

term,  and  what  the  sum  of  the  terms?  Ans.  Z=14,  5=105. 

2.  In  2,  5,  8 of  17  terms,  find  Z and  s. 

Ans.  1—  50,  5= 442. 

3.  In  7,  7J,  74,  - - - - of  16  terms,  find  Z and  s. 

Ans.  Z=  10|,  s=142. 

4.  In  f,  |, of  20  terms,  find  Z and  s. 

Ans.  Z=  — 1-1,  5=  — 13 J. 

5.  In  0,  1,  i, of  11  terms,  find  Z and  s. 

Ans.  1—5,  s=27|. 

6.  In  —10,  -12,-14, of  6 terms,  find  Z and  s. 

Ans.  Z=  — 20,  s=— 90. 

7.  Given  a = 2,  n = 5,  l—  22,  to  find  cl  and  s. 

Ans.  cl=  5,  s=60. 

8.  Given  cl= 2,  « = 12,  5 = 96,  to  find  a and  Z. 

Ans.  « = — 3,  Z=19. 

9.  One  hundred  stones  being  placed  on  the  ground  in  a straight 

line,  two  yards  apart,  how  far  will  a person  travel  who  shall  bring 
them  one  by  one  to  a basket,  placed  at  two  yards  from  the  first 
stone?  Ans.  20,200  yds. 

10.  A railway  train  moves  two  yards  the  first  second,  four  yards 
the  second  second,  six  yards  the  third.  In  how  long  a time  will  the 
train  be  traveling  at  the  rate  of  a mile  a minute? 

Ans.  14.33  sec. 

114.  Ratio. 

Ratio  is  the  relative  magnitude  of  two  quantities  of  the  same 
kind.  The  measure  of  this  relationship,  or,  as  it  is  commonly  said, 
the  ratio  itself,  may  be  always  found  by  dividing  one  of  the  quan- 

b ct 

titles  by  the  other ; thus,  if  a and  Z>  are  the  quantities,  then  — or  - , 

expresses  the  number  of  times  the  one  contains  the  other,  and  is 
their  ratio,  or  the  measure  of  their  relative  magnitudes. 

There  is  some  difference  in  usage  as  to  which  quantity  shall  be 
made  the  divisor ; thus,  the  ratio  of  3 : 6 is  about  as  often  written, 
3 6 

- as  - . The  question  depends  upon  which  of  the  two  is  regarded 
u o 

as  the  standard ; the  ratio  in  this  case  being  4,  if  6 is  taken  as  the 


RATIO  AND  PROPORTION. 


131 


measure ; or  2,  if  3 is  so  taken.  It  is  perhaps  better  to  make  the 
quantity  mentioned  first,  called  the  antecedent , the  divisor,  and  the 
second  quantity,  called  the  consequent,  the  dividend.  At  any  rate, 
we  shall  adopt  this  method. 


115.  Proportion. 

When  two  ratios  are  equal  to  each  other ; as, 

b _d 
a c ’ 


(1) 


the  four  quantities  are  said  to  be  in  proportion.  They  are  often 
written, 

a : b : : c : d. 

This  and  equation  (1)  express  entirely  the  same  truth,  and  the 
one  may  at  any  time  be  used  for  the  other. 

We  may  say,  then,  that, 

A proportion  is  an  equality  of  two  ratios. 

The  ratios  are  called  couplets  ; thus,  a : b is  the  first  couplet; 
c : cl  is  the  second  couplet.  Of  the  four  quantities  in  a proportion, 
the  last  one  is  called  a fourth  proportional  to  the  other  three.  If  the 
second  term  is  used  also  as  the  third,  such  term  is  called  a mean 
proportioned.  In  this  case  the  last  term  is  called  a third  propor- 
tional. The  first  and  last  terms  are  called  extremes  ; the  second  and 
third  are  called  means. 


116.  Let  us  have  the  proportion, 

a : b ::  c : d\ 
or,  writing  it  as  an  equation, 

b_d m 
a c’ 

whence,  be— ad.  - - (1) 

Hence,  we  may  say  that, 

The  product  of  the  means  is  equal  to  the  product  of  the  extremes. 
The  converse  of  this  is  equally  true ; for  dividing  each  member  of 
(1)  by  ac,  we  have, 

b d , y 

- = - ; or  a : b : : c : d. 
a c 


132 


ELEMENTS  OE  ALGEBRA. 


Whence, 

If  the  product  of  two  quantities  is  equal  to  the  product  of  two  other 
quantities,  two  of  them  may  he  made  the  extremes,  and  two  the  means 
of  a proportion. 


117.  Assume  again  the  proportion, 


a : b ::  c : d;  or 


b 

a 


(1) 


We  may  multiply  both  terms  of  each  fraction  by  any  quantity,  as 
m ; thus. 


mb  md  7 

■ — = — ; /.  ma  : mb  ::  me  : md. 
ma  m c 


(2) 


We  may  also  divide  both  terms  of  each  fraction  by  any  quantity, 
as  m ; thus, 

b_  d 

m _m . . a ' b " c ' d . , 

a c ’ ” m ' m m ’ m ' ' 

m m 

We  may  extract  any  root,  as  the  ?«th,  of  both  members  of  (l),and 
shall  have, 

= Va  ■ Vb  : : V c : tyd.  - - (4) 

V a yc 

Or,  again,  we  may  raise  both  members  of  (1)  to  the  ?nth  power, 
thus, 

bm  dm  7 7 

— = /.  am  : b ::  cm  : dm. (o) 

a c 

Hence,  from  (2),  (3),  (4),  and  (5),  it  follows  that, 

1.  We  may  multiply  all  the  terms  of  a proportion  by  the  same  quan- 
tity. 

2.  We  may  divide  all  the  terms  of  a proportion  by  the  same  quan- 
tity. 


JRemark. — It  is  plaia  tliat  the  multiplier  or  divisor  may  he  different  for  each 
couplet. 

3.  We  may  extract  the  same  root  of  every  term  of  a proportion. 

4.  We  may  raise  every  term  to  the  same  power. 


RATIO  AXD  PROPOETION. 


133 


118.  Dividing  unity  by  each  member  of  the  equation 

b cl  . 

- = - ; we  have, 
a c 


b : a : : cl  : c. 


Whence, 

Consequents  may  be  made  antecedents  and  antecedents  consequents. 

The  quantities  are  then  said  to  be  in  proportion  by  inversion. 

Multiplying  both  members  of 

b cl  . . c . 

by  we  have, 


a 
c ' 

d 

V 


a : c ::  b : cl. 


Whence, 

The  antecedents  may  be  made  one  couplet , and  the  consequents 
another.  The  quantities  are  then  said  to  be  in  proportion  by 
alternation. 


119.  Adding  unity  to  both  members  of 


b _d 
a c’ 


and  then  subtracting  unity  from  both,  we  have, 

b d b , cl  , 

-+1=-  + 1; 1 = — 1. 

a c a c 

b+a  d+c  b—a  d—c 


Whence, 


From  which  we  may  write, 

a : b + a ::  c : d + c,  a : b—c  ::  c : d—c. 


Hence,  we  may  say  that, 

The  first  antecedent  may  be  added  to  its  consequent,  provided  the 
second  antecedent  is  added  to  its  consequent.  The  quantities  are  then 
said  to  be  in  proportion  by  composition.  In  the  same  way  the  conse- 
quents may  be  subtracted.  The  quantities  are  then  said  to  be  in 
proportion  by  division. 

Let  us  have  two  proportions  with  a couplet  the  same  in  each ; thus, 
b_cl  b _g 
a~  c’  a-/’ 


134 


ELEMENTS  OF  ALGEBRA. 


Then, 


That  is, 


d _g 

c~f 

c : d ::  f:  g. 


Jf  the  first  couplets  are  the  same  in  two  proportions,  the  other  two 
couplets  form  a proportion. 


„ l)  d mb  nd  ,T  . . 

120.  From  — =-  we  may  write,  — = — . Now,  making  m = 

a c ma  nc 

0)  V . . . 

1±  , and  « = 1±-,  in  this  equation  we  shall  have,  after  a slight 

q s 

transformation, 

b±1--b  d±--d 
q _ s 

p r 

a±--a  c±--c 

q s 

Whence,  b±-- a : b±!--b  ::  c±-- d : d±--d. 

q q s s 

That  is, 

We  may  increase  or  decrease  antecedent  and  consequent  by  like 
parts  of  each. 


121.  Let  us  have  two  proportions, 

a : b ::  c : d ) b d q n 
f : g ::vn:  n ) c f m 

Multiplying  the  equations,  member  by  member, 

by  _ d n 
af  ~ cm’ 

Whence,  af  : bq  : : cm  : dn. 

That  is, 

We  may  multiply  proportions  togeth  er,  term  by  term. 


122.  From  a : b : : c : d, 
we  have  bc—ad\  adding  ab  to  both  members, 

bc  + ab=ad  + ab. 
Whence,  b(a  + c)—a{b  + d). 


GEOMETRICAL  PROGRESSION. 


135 


And,  from  a previous  principle, 

a : b : : a + c : b+d. 

Hence, 

Antecedent  is  to  its  consequent , as  the  sum  of  the  antecedents  is  to 
the  sum  of  the  consequents. 


123.  Where  several  proportions  are  written  together,  thus: 

a : b ::  c : cl  ::  f : g , etc., 

it  is  called  a continued  proportion. 

It  is  sometimes  written  thus, 

a : c : / : : b : d : g. 

The  principles  already  explained  may  be  readily  extended  to  con- 
tinued proportions. 

124.  The  mean  proportional  between  two  quantities  may  be  readily 
found  : thus,  let  it  be  required  to  find  the  mean  between  a and  b. 

We  have,  a : x : : x : b, 

xz=ab, 
x = fab. 

That  is. 

The  mean  of  any  tiuo  quantities  may  be  found  by  multiplying  the 
quantities  together  and  extracting  the  square  root  of  the  product. 

125.  When  one  of  two  variables  is  expressed  in  terms  of  the  recip- 
rocal of  the  other,  as, 

1 m 

x—  - ; or  x—  — , 

V V 

the  quantities  are  said  to  be  reciprocally  proportional.  It  is  manifest 
that  one  will  increase  as  the  other  diminishes,  and -they  are  thus  said 
to  vary  inversely. 

From  this  expression,  we  have, 

xy— 1,  or  xy—m. 

Hence,  the  product  of  two  such  quantities  is  always  constant. 

126.  Geometrical  Progression. 

A Geometrical  Progression  is  a series,  any  term  of  which  may  be 


136 


ELEMENTS  OF  ALGEBRA. 


derived  from  the  one  preceding,  by  multiplying  it  by  a constant 
quantity  called  the  ratio.  It  is  a continued  proportion. 

The  progression  will  be  increasing  when  the  ratio  is  greater  than 
unity:  thus, 

2,  4,  8,  16,  32,  etc., 

is  an  increasing  progression,  whose  ratio  is  2. 

The  progression  will  be  decreasing  when  the  ratio  is  less  than 
unity;  thus, 

32,  16,  8,  4,  2,  1,  i,  etc., 
is  a decreasing  progression,  in  which  | is  the  ratio. 

127.  Formula  for  East  Term. 

Let  us  assume  the  geometrical  progression, 
a : b : c : d : e : f : etc., 
in  which  r is  the  ratio. 

From  the  definition  we  shall  have, 

b=ar,  c=br=ar 2,  d—cr=ar 3,  etc. 

Whence  we  see  that  the  exponent  of  r,  being  unity  in  the  expres- 
sion for  b,  goes  on  increasing  by  unity  for  c,  d,  etc. ; so  that  for  the 
term  which  has  n terms  before  it,  calling  it  1,  we  shall  have, 

l—arn~\ 

Hence,  we  may  say  that, 

Any  term  of  a geometrical  progression  may  be  found  by  multiply- 
ing the  first  term  by  the  ratio  raised  to  a power  whose  exponent  is 
equal  to  the  number  of  preceding  terms. 

128.  Formula  for  the  Sum. 

To  find  a formula  for  computing  the  sum  of  any  number  of  terms 
of  such  a series,  let  us  take  a progression  of  a definite  number  of 
terms, 

a i b : c : d : - - - - : j : k : l, 
the  ratio  being  r. 

Replacing  each  term  after  the  first  by  its  value  in  terms  of  the 
first  term  and  the  ratio,  and  representing  the  sum  of  n terms  by  s, 
we  shall  have, 

s=a+ar+arz  + - - - arn~*  + arn~\ 

How,  multiplying  both  members  of  this  by  r, 

sr—ar  + ar2+ar3+  - - - arn~x  + arn. 


GEOMETRICAL  PROGRESSION. 


137 


Subtracting  the  first  of  these  equations  from  the  second,  we  shall 
have, 


whence, 


sr—s—air—a : 


ar  —a  . . , , _ 

s= — ; or,  since  arn  is  equal  to  Ir, 


s=- 


r — 1 
Ir—a 


r—1  ’ 

an  expression  for  the  sum  of  any  given  number  of  terms. 
We  have  thus  the  two  formulas, 

l—arn~x  and 
Ir—a 


r—1 


, in  which  there  are  five  arbi- 


trary quantities.  If  we  know  all  but  one  which  enter  either  of  them, 
we  can  substitute  the  given  values  in  such  formula,  and  at  once  de- 
duce the  remaining  one;  thus,  if  a— 2,  r= 2,  and  n=5,  we  shall 
have  from  the  first, 

7=2  x 24  = 32. 


If  s=15,  a— 1,  and  r= 2,  from  the  second  formula, 

15  = ^ — ; whence,  1=8. 

Z — 1 


129.  To  find  the  Formula  for  any  Element. 

When  any  three  of  the  five  quantities,  a,  l,  r,  n,  and  s,  which  enter 
these  two  formulas,  are  given,  we  may  combine  the  formulas,  and 
eliminate  one  of  the  remaining  quantities,  and  thus  find  the  fifth 
quantity.  When  a combination  of  the  formulas  is  required,  make 
the  combination  and  deduce  a general  expression  for  the  desired 
quantity,  before  substituting  for  the  given  quantities. 

For  example,  let  s = 62,  r= 2 and  n= 5.  Combiningthe  formulas, 
eliminating  1,  we  have, 


whence,  substituting  the  given  values, 

2°  —I 

Substituting  a= 2,  r= 2 and  n — 5 in  the  first  formula,  we  have, 
l- 32. 

When  n is  to  be  found,  the  resulting  formulas  will  require  the  use 
of  logarithms,  which  are  yet  to  be  explained. 


138 


ELEMENTS  OF  ALGEBRA. 


Examples. 

1.  Given  a= 1,  r=2,  n — 7,  to  find  l and  s.  Ans.  1= G4,  s=127. 

2.  Given  a — 4,  r— 3,  n— 10,  to  find  l and  s. 

Ans.  /=  78732,  s=11809G. 

3.  Given  a= 9,  r=J,  « = 7,  to  find  l and  s. 

Ans.  I— 258£,  s = 591T\ 

4.  Given  « = 6£,  r=\,  n— 8,  to  find  l and  s. 

Ans.  J=106f,  s = 3074. 

5.  Given  a— 8,  r=\,  n — 15,  to  find  l and  s. 

Ans.  ^-g-oVg,  s=15  + . 

6.  Given  a—f,  r— 1|  « = 11,  to  find  / and  s. 

Ans.  ?=xrtfx, 


130.  To  insert  Geometrical  Means. 

Let  it  now  be  required  to  insert  a given  number  of  geometrical 
means  between  any  two  numbers,  as  a and  b.  These  two  quantities, 
together  with  the  means,  will  form  a geometrical  progression  of  two 
more  terms  than  the  number  of  the  means.  If  there  are  to  be  in 
means  there  will  be  in + 2 terms.  Now,  from  the  formula 


This  is  the  value  of  the  ratio  when  there  are  n terms,  and  a and  l 
are  the  extremes.  But  in  the  case  in  hand,  a and  b are  the  extremes 
andm  + 2the  number  of  terms.  Substituting  these  in  the  above 
expression  for  r,  we  shall  have, 


This  being  the  new  ratio,  we  may  now  find  the  means  successively 
from  the  first  term  a ; thus, 


For  example,  let  it  be  required  to  insert  3 geometrical  meaus  be- 
tween 2 and  32. 


In  this  case 


m — 3,  a = 2,  and  b= 32. 


GEOMETRICAL  PROGRESSION. 


139 


These  in  (1)  give,  r=V^=='V/16=2. 

Whence  the  means  will  he  4,  8,  and  16,  and  we  shall  have  the  pro- 
gression, 

2 : 4 : 8 : 16  : 32. 

Examples. 

1.  Insert  5 geometrical  means  between  3 and  192. 

Ans.  3 : 6 : 12  : 24  : 48  : 96  : 192. 

2.  Insert  5 geometrical  means  between  5 and  1215. 

Ans.  5 : 15  : 45  : 135  : 405  : 1215. 


131.  The  Sum  of  an  Infinite  Progression. 

Let  us  now  have  a decreasing  progression  of  an  infinite  number  of 
terms.  The  ratio  will  be  less  than  unity.  The  formula, 

arn—a  ... 

s=— — — , may  be  written, 

arn  a 
S= r 

r— 1 r— 1 

But,  since  r is  less  than  1,  r 2 will  be  less  than  r,  r3  still  less,  and 
so  the  results  will  go  on  diminishing  as  the  power  is  increased;  and 
since  n is  infinite,  rn  is  zero;  whence  it  follows,  that  the  first  term 
of  the  second  member  of  the  above  expression  is  zero,  and  we  have 

a 


But,  since  r<l,  we  may  write 

a 

S=l-? 

That  is  to  say, 

The  sum  of  the  terms  of  a decreasing  geometrical  'progression  of  an 
infinite  number  of  terms,  is  equal  to  the  first  term,  divided  by  unity 
minus  the  ratio. 

What  is  the  sum  of  the  following? 

1.  1 : £ : i - - - - to  infinity.  Ans.  2. 

2.  1 : J : -$•  - - - - to  infinity.  Ans.  f. 

3.  4:2:1  - - - - to  infinity.  Ans.  8. 


140 


ELEMENTS  OF  ALGEBRA. 


4-  i : is  '•  tIt  - - - to  infinity.  Ans.  \. 

5-  1 : - : \ - - - to  infinity.  Ans.  — — . 

a a2  J a — l 

If  the  ratio  of  a geometrical  progression  is  negative,  the  terms 
will  be  alternately  positive  and  negative.  In  substituting  in  the 
formulas,  be  careful  to  give  the  ratio  its  proper  sign. 

G.  Given,  a=—2,  r—  — 3n—5,  to  find  l and  s. 

Ans.  I—  — 162,  s = — 121. 


SECTION  X. 

LOGARITHMS. 

132.  If  the  number  10  be  raised  to  the  second  power,  we  shall  have 
100  for  the  result;  thus, 

102  = 100. 

The  exponent  2 is  called  the  logarithm  of  100.  Let  us  write 
several  exact  powers  of  10,  thus, 

(10)~2,  (10)-1,  10°,  101,  102,  103,  etc. 

Or, 

aV)2  1,  10,  100,  1000,  etc. 

The  exponents, 

—2,  —1,  0,  1,  2,  3,  etc., 
in  the  upper  row,  are  the  respective  logarithms  of 
.01,  .1,  1,  10,  100,  1000,  etc., 

in  the  lower  row. 

If  we  should  take  any  number  which  is  not  an  exact  power  of  10, 
the  exponent  would  not  be  a whole  number,  but  would  be  made  up 
of  an  entire  part  and  a fraction  ; thus,  the  number  25  is  greater  than 
the  first  power  of  10,  and  less  than  the  second  powTer ; whence  10 
must  be  raised  to  a power  greater  than  1,  and  less  than  2,  to  produce 
25.  Assuming  that  the  fraction  to  be  added  to  1 to  give  the  proper 
exponent  is  .397940,  we  should  have, 

(10)’-3,7940=25. 

Here  1.397940  is  the  logarithm  of  25. 


L0GABITH1T3. 


141 


It  is  obvious  that  any  number  except  unity  may  be  used  instead 
of  10,  as  the  fixed  number,  whose  several  powers  are  to  be  taken. 

We  may,  then,  say  that. 

The  logarithm  of  a number  is  the  exponent  of  the  paver  to  which  it 
is  necessary  to  raise  a fixed  number,  called  the  base,  in  order  to  pro- 
duce the  given  number. 

The  base  of  the  Common  System,  as  it  is  generally  called,  is  10. 

The  base  of  the  Napier  an  System,  so  named  from  Baron  Napier, 
to  whom  the  invention  of  logarithms  is  due,  is  2.71828182.... 

The  entire  part  of  a logarithm  is  called  the  characteristic  ; the  deci- 
mal part  is  called  the  mantissa. 


133.  Characteristic. 


The  characteristic  of  the  logarithm  of  a number  may  always  be 
written  at  once  from  the  number  itself.  The  law  for  writing  the 
characteristic  may  be  discovered  from  the  following : 


(10)4  =1,0000 
(10)3  = 1,000 

(10)2  = 100 

(10)1  = 10 

(10)°  = 1 

(10)-1  =.l 
(10)-2  =.01 
(10)-3  =.001 
(10)-4  =.0001 


log 

10,000  = 

4. 

log 

1,000  = 

n 

O. 

log 

100  = 

2. 

log 

10  = 

1. 

log 

1 = 

0, 

log 

.1  =- 

-1. 

log 

.01  fj 

-2. 

log 

.001  =- 

-3. 

log 

.0001  =- 

-4. 

It  thus  appears  that  the  characteristic  of  all  numbers  greater  than 
unity  is  positive,  and  that  the  characteristic  of  all  numbers  less  than 
unity  is  negative  ; and  that, 

When  the  number  is  greater  than  unity,  the  characteristic  is  one 
less  than  the  number  of  digits  composing  the  number.  If  the  num- 
ber is  partly  decimal,  only  the  entire  part  is  to  be  counted. 

When  the  number  is  less  than  unity,  and  is  expressed  as  a deci- 
mal, the  characteristic  is  negative  and  one  greater  than  the  number 
of  0’s  immediately  following  the  decimal  point. 

The  mantissa,  or  decimal  part,  must  be  found  in  a table  computed 
for  the  purpose. 


134.  The  following  is  the  beginning  of  such  a table,  showing  the 
logarithms  from  1 to  100  : 


142 


ELEMENTS  OP  ALGEBRA. 


TABLE. — Logarithms  from  1 to  100. 


N. 

Log. 

N. 

Log. 

N. 

Log. 

N. 

Log. 

1 

0 000000 

26 

1 414973 

51 

1 707570 

76 

1 880814 

2 

0 301030 

27 

1 431364 

52 

1 716003 

77 

1 886491 

3 

0 477121 

28 

1 447158 

53 

1 724276 

78 

1 892095 

4 

0 602060 

29 

1 462398 

54 

1 732394 

79 

1 897627 

5 

0 698970 

30 

1 477121 

55 

1 740363 

80 

1 903090 

6 

0 778151 

31 

1 491362 

56 

1 748188 

81 

1 908485 

7 

0 845098 

32 

1 505150 

57 

1 755875 

82 

1 913814 

8 

0 903090 

33 

1 518514 

58 

1 763428 

83 

1 910078 

9 

0 954243 

34 

1 531479 

59 

1 770852 

84 

1 924279 

10 

1 000000 

35 

1 544068 

60 

1 778151 

85 

1 929419 

11 

1 041393 

36 

1 556303 

61 

1 785330 

86 

1 934498 

12 

1 079181 

37 

1 568202 

62 

1 792392 

87 

1 939519 

13 

1 113943 

38 

1 579784 

63 

1 799341 

88 

1 944483 

14 

1 146128 

39 

1 591065 

64 

1 806180 

89 

1 949390 

15 

1 176091 

40 

1 602060 

65 

1 812913 

90 

1 954243 

16 

1 204120 

41 

1 612784 

66 

1 819544 

91 

1 959041 

17 

1 230449 

42 

1 623249 

67 

1 826075 

92 

1 963788 

18 

1 255273 

43 

1 633468 

68 

1 832509 

93 

1 968483 

19 

1 278754 

44 

1 643453 

69 

1 838849 

94 

1 973128 

20 

1 301030* 

45 

1 653213 

70 

1 845098 

95 

1 977724 

21 

1 322219 

46 

1 662758 

71 

1 851258 

96 

1 982271 

22 

1 342423 

47 

1 672098 

72 

1 857333 

97 

1 986772 

23 

1 361728 

48 

1 681241 

73 

1 863323 

98 

1 991226 

24 

1 380211 

49 

1 690196 

74 

1 869232 

99 

1 995635 

25 

1 397940 

50 

1 698970 

75 

1 875061 

100 

2 000000 

135.  General  Principles  of  Logarithms. 

The  principles  already  established,  Art.  50,  with  regard  to  expo- 
nents in  general,  govern  the  use  of  logarithms,  they  being  but  a 
class  of  exponents. 

Let  us  repeat  the  principles  referred  to,  however,  in  this  connec- 
tion, since  they  form  the  basis  of  all  logarithmic  computations. 

Let  m and  n be  any  two  numbers  whose  logarithms  are  x and  y. 
Then, 

(10)I  = H1  - - (1) 

(10 /=n  - - (2). 

Multiplying  member  by  member,  we  have, 

(10)I+!,= mn, 

whence,  x+y=  log  mn, 

That  is, 

1.  The  suih  of  the  logarithms  of  two  numbers  is  the  logarithm  of 
their  product. 


LOGARITHMS. 


143 


Again,  from  (1)  and  (2)  we  have, 

(10)"  __mm 
(10)*  ~ n’  °r’ 

VI 

(10)*"*=— ; whence, 
n 


* 


, m 

x~y=  log-« 

That  is, 

2.  The  difference  of  the  logarithms  of  two  numbers  is  the  logarithm 
of  their  quotient. 

We  have  again,  from  (1), 

((10)')?rr»iP 

mpx=mp 
px=  log  Vlp. 

That  is, 

3.  The  product  of  the  logarithm  of  a number  by  p is  the  logarithm 
of  the  p power  of  the  number. 

If  we  extract  the  r root  of  both  members  of  (1),  we  have, 


(10)r  = ffm  :. 

rQ  r 

—A:  log  V Ol. 
r * 

That  is, 

4.  The  quotient  of  the  logarithm  of  a number  by  r,  is  the  logarithm 
of  the  r root  of  the  number. 

These  four  truths  are  called  the  general  principles  of  logarithms. 

Apply  the  foregoing  principles  in  the  transformation  of  the  follow- 
ing expressions  into  corresponding  logarithm  equations: 

Ans.  log  a;=log  ft  + log  b— log  c— log  d. 

A ns.  log  x—m  log  « + « log  b + p log  c. 

Ans.  log  x=m  log  a — n log  b—p  log  c—q  log  d. 


1.  x— 


ab 


cd’ 

2.  x—anbncp. 
amb ~n 


o.  x= 


cpdq  ' 


4.  x—anb  i. 


Ans.  log  x—  — log  a—  - log  b. 


144 


ELEMENTS  OF  ALGEBRA. 


5.  x = y// amb~nci 

a^y cT 


6.  x— 


Ans.  log  x=  — log  a— log  b + — log  c. 
° n ° nq 


m 


7.  x=s 


A?is.  log  cr=log  a-\ — log  c— log  b—\  log  d. 

bVd  n 

{a  + b)ncA 


( c + d ) V d 3 

Ans.  log  x—n  log  {a  + b)+m  log  c— log  {c  + d)  — f-  log  d. 


8.  x— 


d/a  + b 


9.  x—  \/a2—x2. 
10.  x={am)n. 


Ans.  log  x—  — log  {a  + l). 


Ans.  log  a=^-log  {a  + x)+  ^ log  {ci—x). 


Ans.  log  x—nin  log  a. 


136.  Solution  of  Equations. 

By  the  aid  of  logarithms  we  are  enabled  to  solve  what  would  other- 
wise be  difficult  equations. 

Solve  the  following : 


( x log  5= log  25. 

2.  7 V*  = 17. 

Whence,  Vo" log  7 = log  17, 
log  17 

which  gives,  Vx=tyYm 


3.  ax  = b. 


4. 


log  25 
Ans.  x—  W — — • 
log  o 


Ans. 


log  b 
log  a 


Ans. 


( log  g V 

\log  {a  + b)J 


x 

5.  am  — b. 


Ans. 


x — rn 


137.  Logarithms  of  Decimals. 

When  a number  is  composed  of  the  same  figures,  no  matter  how 
its  value  may  be  made  to  change  by  moving  the  decimal  point  to 


LOGARITHMS. 


145 


the  right  or  left,  the  mantissa  will  always  remain  the  same ; the 
only  change  being  in  the  characteristic : thus,  if  the  logarithm  of 
2975376  is  6.354189,  we  shall  have  as  follows : 

log  2975.376  =3.354189. 
log  29.75376  =1.354189. 
log  2.975376  =0.354189. 
log  .02975376  = 2.354189. 

The  mantissa,  it  will  be  observed,  remains  the  same. 

To  prove  that  this  is  true,  let  n be  any  number  whatever,  and  p 
any  other  whole  number,  positive  or  negative. 

Then,  n x (lO)71,  will  represent  the  product  of  n by  any  exact 
power  of  10.  We  may  write, 

log  (»x  (10)7>)=log  ?j  + log  10r=log  n+p. 

Now,  in  the  expression,  log  n+p,  since  p is  entire,  it  will  be  added 
to,  or  if  negative  subtracted  from,  the  characteristic;  so  that  the 
decimal  part  will  remain  intact.  But  to  multiply  a number  by  any 
entire  power  of  10,  positive  or  negative,  is  but  to  remove  the  decimal 
point  to  the  right  or  left ; and  hence  it  follows  that  the  mantissa  of 
the  logarithm  of  a number  expressed  by  any  combination  of  figures 
will  remain  the  same,  whether  any  part  of  it  is  decimal  or  not. 

This  principle  enables  us  to  disregard  the  decimal  point  entirely  in 
finding  the  mantissa  of  the  logarithms  of  numbers. 

138.  General  Properties. 

When  the  base  of  a system  of  logarithms  is  greater  than  unity, 
the  logarithm  of  oo  is  +co,  and  the  logarithm  of  0 is  —go  . 

For  assuming  the  exponential  equation, 

10  x=n, 

n being  any  number  and  x its  logarithm,  we  see  that  as  n increases 
x must  increase,  and  when  n is  infinite  x must  be  infinite  also. 

As  n is  made  less  and  less,  x must  be  diminished,  until  when  n is 
1,  x is  0.  If,  now,  n is  still  diminished,  x becomes  negative,  and 
when  n is  0,  x is  again  co,  and  becoming  so  from  the  negative  side, 
is  still  regarded  as  negative.  Thus,  the  logarithms  of  positive  num- 
bers alone  require  for  their  logarithms  all  numbers  from  — go  to 
+ co.  There  cannot,  then,  be  any  logarithms  of  negative  quantities. 

If  the  base  is  less  than  unity,  the  same  thing  will  hold;  except 
IC 


146 


ELEMENTS  OF  ALGEBRA. 


that  the  logarithms  of  numbers  greater  than  unity  will  then  be  nega- 
tive, and  those  of  numbers  less  than  unity  will  be  positive. 

It  will  be  seen  that  the  logarithm  of  unity  in  any  system  is  zero. 

139.  Moduius. 

Let  us  assume  what  is  called  the  Logarithmic  Series,  leaving  its 
deduction  for  a more  advanced  course.  It  is, 

?/2  yS  y4: 

l°g(1+y)=m(y-'.r  + j-^r+  - - - - etc.), 

in  which  we  have  the  logarithm  of  a number,  (1  + y),  developed  into 
a series,  in  terms  of  a number,  y,  less  by  unity  than  the  number 
itself. 

It  will  be  observed  that  the  second  member  is  composed  of  two 
factors,  m and  the  series  within  the  brackets.  The  factor  m depends 
for  its  value  entirely  upon  the  base  of  the  system,  and,  for  the  same 
system,  is  constant.  It  is  called  the  modulus  of  the  system.  The 
modulus,  then,  of  any  system  of  logarithms  is  the  constant  factor, 
which  is  common  to  all  logarithms  of  that  system. 

The  modulus  of  the  N apieran  system  is  1 ; the  modulus  of  the 
common  system  is  0.43429448. 

140.  Indeterminate  and  Identical  Equations. 

An  Indeterminate  quantity  is  one  which  has  no  fixed  value,  but 
admits  of  an  infinite  number  of  values  in  succession.  Indeterminate 
and  arbitrary  quantities  do  not  differ  in  nature  ; but  it  is  usual  to 
call  such  quantities,  when  represented  by  the  first  letters  of  the 
alphabet,  arbitrary ; and  when  represented  by  the  final  letters,  inde- 
terminate. 

An  Indeterminate  Equation  is  one  which  contains  at  least  two 
quantities  which  can  only  be  assumed  in  relation  to  each  other ; 
thus, 

bx — 4y=9, 

is  such  an  equation  ; x and  y admitting  of  any  number  of  sets  of 
values,  but  neither  of  them  admitting  of  any  specific  value  indepen- 
dently of  the  other.  Such  quantities  are  called  indeterminate,  or, 
perhaps  more  correctly,  variables. 

An  Identical  Equation  is  one  which  is  true  for  all  possible  values 
of  the  indeterminate,  or  arbitrary  quantities,  which  enter  it.  For 
example, 


LOGARITHMS. 


147 


ax  + by—ax  + by, 

(a+x)~  = a2  +2  ax  + x2, 

5x  -t-7a;2  +4=5x  + 7.'C2  +4, 
are  manifestly  sucli  equations. 

An  identical  equation  differs  materially  from  the  ordinary  or  de- 
terminate equation,  which  admits  of  but  a specific  number  of  values; 
and  also  from  the  indeterminate  equation,  which  is  satisfied  for  any 
number  of  sets  of  relative  values  of  the  variables  which  enter  it.  In 
an  identical  equation,  the  co-efficients  of  the  indeterminate  quanti- 
ties (x,  y,  z,  etc.),  are  no  longer  arbitrary,  but  have  resultant  values, 
and  so  are  themselves  determinate. 


141.  Indeterminate  Co-efficients. 

Any  identical  equation  containing  but  one  indeterminate  quantity 
can  be  reduced  to  the  form  of 

a + bx  + cx2  + dx3  + etc.=0  - - (1), 

an  equation  in  which  a,  b,  c,  cl,  etc.,  are  called  Indeterminate  Co-effi- 
cients, since  they  are  co-efficients  of  the  indeterminate  quantity. 

Now,  since  this  equation  is  true  whatever  value  x may  have,  it  will 
be  true  when  x—0.  Making  this  hypothesis  in  the  equation,  we 
shall  have, 

o=0. 

This  value  of  a in  (1)  gives, 

bx  + cx 2 + dx3  +etc.=0. 

Factoring, 

x(b  + cx  + dx2  +etc.)=0. 

This  may  be  satisfied  for  x=0,  or, 

b + cx  + dx2  +etc.  = 0 - - (2) 

Now,  x,  in  this  equation,  must  admit  of  the  same  value  as  in  (1); 
hence,  this  must  also  be  an  identical  equation,  and  is  true  for  x—0. 
This  value  of  x in  (2)  gives, 

b—0 ; 

and  in  the  same  way  we  may  prove, 

c= 0, 
d— 0; 

and  so  for  all  the  co-efficients  of  x. 

Hence,  we  may  say  that, 


148 


ELEMENTS  OF  ALGEBEA. 


In  cm  identical  equation  of  one  indeterminate  quantity,  the  co-effi- 
cients of  that  quantity  are  severally  equal  to  zero. 


143.  Principle  of  Indeterminate  Co-efficients. 

Let  us  now  have  the  identical  equation, 

a + bx  + ex*  -\- Qtc.=za'  + b' x + c x*  +etc. 

Transposing  and  factoring, 

a—a'  + (b—b')x  + (c—c')xz  +etc.=0. 

Now,  from  the  principle  just  deduced, 

a— a'=0  .•.  a— a! 
b-b'= 0 b-V 

c—c'= 0 c—c' 

etc.  etc. 

Whence,  we  may  say  that. 

In  an  identical  equation  of  one  indeterminate  quantity,  the  co-effi- 
cients of  the  like  quivers  of  that  quantity  in  the  two  members  are  sev- 
erally equal  to  each  other. 

This  principle  may  be  readily  extended  to  identical  equations  con- 
taining any  number  of  indeterminate  quantities.  It  is  called  the 
Principle  of  Indeterminate  Co-efficients. 


143.  Development  of  Expressions. 

Let  us  now  apply  the  principle  of  Indeterminate  Co-efficients  to  the 
development  of  algebraic  expressions  into  series. 

Take  the  fraction,  — r,  and  placing  it  equal  to  a development 

1 — X — X“ 

of  the  proposed  form  : we  have, 


"I  -4-  9 p 

— — a + bx+cxz  +dx*  + etc.  - (1). 

l—x—x- 


Since  this  development  must  be  true  for  all  values  of  x,  this 


must  be  an  identical  equation. 

Clearing  it  of  fractions,  using  the  vertical  bar  for  convenience, 
we  have, 


1 + 2 x=a  + b 

x + c 

x ~ + d 

—a 

-b 

— c 

—a 

-b 

x3  + etc. 
+ etc. 
+ etc. 


Whence,  by  the  principle  of  Indeterminate  Co-efficients,  we  shall 
have. 


LOGARITHMS. 


149 


l— a :.  «=1 

2=b—a  .\  b—  3 

0 —c—b—a  :.  c= 4 
0 =d—c—b  :.  d—5 
etc.  etc. 

These  values,  substituted  for  a,  b,  c,  d,  etc.,  in  (1)  give, 
l + 2x 


1—x—x2 
the  desired  development. 


= 1 + 3x  + 4x2  + 5a;3  + etc., 


Examples. 

1 | 

1.  Develop  - — — into  a series.  Ans.  l + 5.T  + 15a;2  + 45.t3  +etc. 


2 ft 

2.  Develop  — — : — - into  a series. 


1 + x + x2 


Ans.  l—2x  + x2+x3  — 2xi+eto. 


3.  Develop into  a series. 

a + x 


x x 


Ans.  1— 1 + — — — r + etc. 


a a* 


2 2) 

4.  Develop  - — — — into  a series. 

1 — 2x— 3x2 


Ans.  l + .-r  + Sx3 +13a;3 +etc. 


5.  Develop  - 


2x  + x2 
In  this  case  we  have, 


into  a series. 


"Whence, 


=a  + bx  + cx2  +etc. 

2x  + x2 

x3  +etc 
+ etc. 

1 = 0 
0=2  a 
etc. 


1 = 2 a 

x + 2 It 

x2  +2  c 

+ • a 

+ b 

Here  we  encounter  an  absurdity  in  the  result,  1=0. 

This  shows  that  the  development  cannot  be  made  as  proposed. 

Then,  let  us  factor  the  expression  thus,  - x - : now  let  us  de- 

CC  /v  “p  Oj 


velop 


2 + x 


. We  have, 


2 + x 


-~a  + bx  + cx  + etc. 


150 


ELEMENTS  OF  ALGEBRA. 


Proceeding  as  usual,  we  have, 


And  thus  we  have, 


l=2a  + 2b\x  + 2c\x  + etc. 

a\  b\  +etc. 
1=2 a.  a — 

0=2 b + ci.  l — —\. 

0=2c  +b.  c— 

etc.,  etc. 


1 _1 

2 ~i~  x 2 


1 1 „ 

~±x  + %x  +etc. 


Now,  multiplying  both  members  by  — , we  have, 


2 x + x2  2x  4 8 


5i.-r+sa;+etc5 


which  is  the  true  development  of  the  original  expression. 

1 . 

Here  the  first  term  — is  the  same  as  1— - ; so  that  the  absurditv 
2x  2 ' 


above  arose  from  not  starting  our  development  with  a low  enough 
power  of  x. 

So,  in  general,  when  such  an  absurdity  develops  itself,  it  will  be 
found  due  to  a like  cause. 


143.  Partial  Fractions. 


The  Principle  of  Indeterminate  Co-efficients  affords  an  easy  method 
of  resolving  a fraction  into  its  partial  fractions. 

To  do  this,  place  the  given  fraction  equal  to  the  sum  of  as  many 
partial  fractions  as  the  denominator  of  the  fraction  has  factors,  the 
several  numerators  of  such  assumed  partial  fractions  being  p,  q,  r, 
etc. — quantities  to  be  determined — and  the  denominators  being  the 
several  separate  factors  of  the  denominator. 

For  example,  let  it  be  required  to  find  the  partial  of  — — 

We  have, 

- — =— — | — — , an  identical  equation. 
a2—xz  a + x a—x 


Clearing  of  fractions, 
Whence, 


2 a = ap  —px  + aq  + qx. 

2 a=  ap  + aq; 

0 =—p  + q. 


Combining  these  two  equations  and  eliminating  q,  we  have, 

P-1- 


This  value  of  p gives, 
Therefore, 


LOGARITHMS. 


151 


2=1- 

2a  _ 1 1 

a2 — x2  a + x a — x" 


Examples. 

3^ 3 2 2 

1.  Eesolve  — — - into  partial  fractions.  Ans.  „4 — — -. 

x— 3 x+6 


2.  Resolve 


a;3 -9 

x2  + 4x— 11 
(x—  1 ) (x  + 2)  (x—3) 


into  partial  fractions. 

, 1 1 1 

Ans.  — H 

x — 1 x + 2 X — a 


1 ' ^iCC  | 'X/  ^ 

3.  Resolve  — .1  ■ into  partial  fractions. 


(1— a;)3 

When  there  are  equal  factors  in  the  denominator,  as  in  this  case, 
use  all  of  the  several  powers  ; thus, 

1 — 2x  + x2 


V 


* - +:  ' 


(1— a;)3  (1— a:)3  (1— a;)3  1— a;’ 

i 

Ans. 


1 1 

+ : 


1— a;  1 — a:  1— a; 


4.  Resolve 


a + x 


5.  Resolve  - 


(a  + x 
-16 


— into  partial  fractions.  Ans.  - 
•)3  1 « + ■ 


x a + x 


x 3 + 2x— 15 


into  partial  fractions. 


, 2 2 

Ans.  . 

a;  + o x—o 


NOTE. 

PROBLEM  OF  THE  LIGHTS. 


Tiie  point  maintained  in  the  discussion  of  the  problem  of  the  lights,  that  c= 
0,  when  a is  greater  or  less  than  b,  is  a legitimate  hypothesis,  and  that  the  an- 
alytical result  is  to  be  interpreted  in  a strict  and  natural  way,  is  so  entirely  at 
issue  with  all  the  late  writers  upon  algebra,  that  a more  detailed  examination 
of  the  question  seems  to  be  demanded  than  could  well  be  given  it  in  the  body  of 
the  text. 

Let  us,  then,  consider,  first,  the  arguments  upon  which  we  are  asked  to  aban- 
don the  analysis  in  favor  of  what  is  supposed  to  be  common-sense. 

It  will  be  remembered  that  the  roots  of  the  equation  derived  from  the  con- 


ditions of  the  problem  are,  x=- 


Va 


. Now,  when  c is  made  equal  to  zero 

Va  + Vb 

in  these  roots,  rtand&  remaining  unequal,  we  must  have  x—0,  for  the  value  of 
either  root.  The  interpretation  of  this  result,  according  to  the  general  law, 
gives  us  the  origin  of  distances  as  the  single  point  of  equal  illumination.  We 
then  have  the  case  of  two  lights,  placed  at  the  same  point,  differing,  however, 
much  in  intensity,  and  are  required,  by  the  analysis,  to  conclude  that  such  point 
is  equally  illuminated  by  the  greater  and  the  lesser  light.  This  conclusion,  we 
are  told,  is  not  true. 

The  first  distinguished  author,  whose  reasoning  upon  the  point  we  shall  ex- 
amine, says,  in  his  University  Algebra,  “It  is  obvious  that  when  two  lights, 
of  unequal  intensities,  occupy  the  same  place,  there  is  no  point  in  space  equally 
illuminated  by  them  ; not  even  the  point  in  which  they  are  both  situated.” 

To  show  how  the  result,  x=0,  in  this  case  fails,  this  author  carries  us  back  to 

the  original  equation,  — — p>  1 which,  we  agree  with  him,  “truly  repre- 

sents the  conditions  of  the  problem.” 
a _b 
x 4 x * 

a>b  or  a < b.  For,  by  substituting  any  value  for  x,  ve  shall  always  obtain  two 
unequal  fractions.  If  *=0,  the  two  members  are  two  unequal  infinities.”  And 
so  he  concludes,  that,  under  this  hypothesis,  “ the  problem  fails  altogether,  and 
is  impossible.”  But  can  two  infinite  distances  be  unequal  ? It  is  to  be  presumed 
that  no  one  will  dispute  that  an  infinite  length  is  a distance  than  which  nothing 
can  be  conceived  to  be  greater.  But  to  say  that  two  distances  are  unequal  in 
length,  is  incontestably  to  limit  the  less,  and  so  to  reduce  it  at  once  to  the  finite  ; 
hence,  one  of  this  author’s  infinities  must  be  finite — a contradiction  in  terms. 
The  argument  failing,  the  analytical  deduction,  £=0,  still  demands  our  con- 
fidence. 

But,  if  possible,  a more  surprising  turn,  to  avoid  the  difficulty,  is  made  by 
another  eminent  authority.  It  is  the  more  marked,  too,  in  this  case,  from  the 
fact  that  there  has  been  a deliberate  departure  in  the  last  edition  of  this  author's 


He  says,  “ If  we  put  e=0,  the  result  is 
an  equation  which  cannot  be  satisfied  by  any  value  of  x whatever,  while 


PROBLEM  OF  THE  LIGHTS. 


153 


University  Algebra,  as  well  as  in  liis  more  elaborate  work — a work  which  has 
long  held  its  place  as  a standard  authority.  In  the  previous  editions  the  result 
of  the  analysis  upon  this  point  was  accepted  as  true,  but  without  any  attempt 
at  an  explanation  of  the  difficulty.  We  are  now,  however,  told  that 

“ The  hypothesis  of  e— 0 places  the  two  lights  at  the  same  point,  in  which 
case  they  form  one  and  the  same  light,  whose  intensity  is  equal  to  the  sum  of 
their  intensities  taken  separately.  The  conditions  of  the  problem  involve  the 
necessity  of  two  lights,  and  the  equation  of  the  problem  is  found  under  this 
hypothesis.  This  equation  ought  not,  therefore,  to  respond  to  the  case  of  a 
single  light.  For,  the  interpretation  of  the  result  obtained  from  one  equation, 
can  only  give  the  cases  which  fall  under  the  hypothesis.  The  hypothesis  of 
two  lights,  and  the  hypothesis  of  a single  light,  are  not  connected  by  any  law 
which  affords  a common  equation  of  condition.  Hence,  the  results  obtained  on 
the  supposition  of  c=0,  do  not  belong  to  the  problem.” 

Now  this  reasoning  has  a very  plausible  look  ; but  it  should  seem  that  it  will 
not  bear  investigation.  In  the  first  place,  the  two  lights  do  not  become  one  in 
the  writer’s  sense,  any  more  than  two  circles  of  unequal  radii  become  one  when 
they  have  a common  centre.  Again,  if  one  of  the  lights  be  removed,  the  other 
will  remain,  and  since  there  is  nothing  requiring  the  lights  to  act  simulta- 
neously, the  real  question  in  issue  is  to  find  the  point  or  points  which  would  be 
equally  illuminated  by  either  light,  separately,  and,  if  you  please,  at  times,  how- 
ever remote  from  each  other. 

Again,  there  is  a most  unmistakable  connection  between  the  two  lights — the 
very  arbitrary  quantity  c in  question.  Suppose  it  should  be  required  to  find  the 
locus  of  a point  moving  so  that  its  distance  from  two  fixed  points  shall  be  equal 
to  a given  line.  Here  are  two  points,  but  shall  we  be  told  that  the  distance  be- 
tween them  may  not  be  made  zero,  and  so  give  us  the  circle — a particular  case 
of  the  ellipse  ? or  that,  having  the  equation  of  a line  passing  through  two  points, 
the  two  points  may  not  be  made  one  without  destroying  the  equation  ? 

Again,  this  author  tells  us,  in  the  works  in  question,  that  the  discussion  of  a 
problem  consists  in  making  every  possible  supposition  upon  the  arbitrary 
quantities  which  enter  it,  and  interpreting  the  results.  How  does  this  agree 
with  his  assertion  in  this  case,  that  the  “results  obtained  by  making  c=0  do  not 
belong  to  the  problem”  ? Are  we  to  understand  that  he  considers  it  impossible  to 
make  this  hypothesis  in  an  equation  containing  c ? Indeed  the  doctrine  here 
put  forth  would  make  short  work  with  much  of  the  higher  mathematics,  and 
the  whole  doctrine  of  limits. 

Nor  would  the  difficulty  in  point  be  at  all  obviated,  if  one  were  to  admit  the 
whole  of  this  writer’s  reasoning.  The  equation  must  still  respond,  so  long  as  c 
is  not  absolutely  zero.  Well,  now  suppose  the  lights  to  be,  the  one  very  great 
and  the  other  very  small,  and  that  c is  an  infinitesimal.  We  shall  have  the 
lesser  light  occupying  the  second  or  third  consecutive  point  from  the  greater. 
We  must  have  one  point  of  equal  illumination  between  them  ; but  will  common 
sense  with  regard  to  lights,  which  seems  to  frighten  our  author  into  deserting 
the  analysis,  be  in  any  manner  better  saved  ? Will  not  the  feebler  one  be  hope- 
lessly outshone  by  the  more  splendid  luminary  ? 

But  a third,  and  the  latest  writer  upon  this  point,  pronounces  the  results  of  the 
analysis  under  the  hypothesis  in  question,  to  be  “ evidently  absurd.”  Tie  says, 
“ In  discussing  this  problem,  some  have  committed  the  error  of  considering  that, 


154 


ELEMENTS  OF  ALGEBRA. 


since  for  c=0  and  a and  6 unequal,  x=c  - 


— =0,  therefore,  there  is  a point 

Ya±  Vb 

of  equal  illumination  at  the  point  where  the  lights  are  situated  ! This  is 
evidently  absurd,  since  the  hypothesis  is  that  the  lights  are  of  unequal  inten- 
sity. The  error  consists  in  not  perceiving  that  the  hypothesis,  c=0,  excludes 
the  hypothesis,  a and  b unequal.  That  the  hypotheses,  a>  or  <b,  are  excluded  by 
the  hypothesis,  e=0,  and  that  there  is  a point  of  equal  illumination,  is  self- 
evident.” 

It  is  to  be  presumed  that  the  learned  author  uses  the  word  “ self-evident”  in 
an  unusual  sense,  since  the  question  he  disposes  of  so  summarily  in  the  end, 
had  already  cost  him  quite  an  argument,  to  say  nothing  of  having  been  a stand- 
ing puzzle  for  many  years.  This  conclusion  is  further  sustained  by  the  fact 
that  he  afterwards  says,  “ Perhaps  the  student  may  think  that  these  conditions 
are  no  more  inconsistent  than  those  in  I.  3 above,  viz.,  c finite  and  a=b,  and  a 
point  of  equal  illumination,”  etc. 

He  goes  on  further  to  prove  his  “ self-evident  ” proposition  as  follows : 


a 


Also,  that  a > or  < b is  inconsistent  with  c=0  and  — = 


(that  is. 


x*  (c—x)2 

that  there  is  a point  of  equal  illumination),  appears  from  the  fact  that  c=0 

renders  the  latter  ~ , whence  a—b.” 

x 2 x 2 

This  writer  seems  to  be  unconscious  of  the  fact  that  he  has  here  stricken  out 
zero  as  a factor  from  both  members  of  the  equation  ; thus,  ax2=bx2,  a-0=6  0 .'. 
a=b  ; by  which  process  it  is  easy  enough  to  prove  2=4 ; a result  which  follows 
at  once,  if  we  assume  a= 2,  and  6=4,  in  the  above  equation.  It  should  seem 
that  the  anxiety  to  escape  a point  which  is  thought  to  militate  against  common 
sense,  has  led  to  some  hasty  writing. 

And  now  let  us  look  at  the  question  upon  its  own  merits.  It  should  seem 
that  the  chief  difficulty  arises  from  a most  unwarrantable  mixing  up  of  physical 
phenomena  and  analytical  exactness.  Analysis,  per  se,  knows  nothing  what- 
ever of  the  physical  world  ; nor  does  it  in  the  slightest  degree  regard  any  lan- 
guage but  its  own.  It  takes  our  conditions  and  renders  its  verdict,  leaving  us 
to  look  out  for  ourselves  and  our  physical  applications. 

The  confusion  in  the  minds  of  those  authors  who  desert  the  analysis  in  this 
question,  and  betake  themselves  to  what  they  mistake  for  common  sense,  arises 
from  not  sufficiently  regarding  the  assumed  law  of  physics  under  which  the 
equation  in  this  problem  is  established.  These  are  purely  theoretical  lights, 
and  the  law  is  absolute,  that  their  intensity  shall  vary  inversely  as  the  square  of 
the  distance.  When  we  pass  within  the  distance  unity,  their  intensities  increase 

with  wondrous  rapidity,  until  at  zero — the  limit  of  approach — we  must  have  — , 


for  their  respective  intensities : that  is  to  say,  the  intensity  of  either 


light  at  this  point  is  absolutely  infinite  ; and  thus,  unless  we  may  indeed  have 
“ two  unequal  infinities,”  when  the  lights  occupy  the  same  point,  that  point 
must  be  equally  illuminated,  however  much  the  lights  may  differ  in  intensities 
at  the  distance  unity. 

This  may  not,  probably  does  not,  accord  with  physical  facts,  which  deal  with 


PROBLEM  OP  THE  LIGHTS. 


155 


tallow  candles,  or  at  best  calcium  lights  ; but  let  us  not  charge  the  failure  upon 
the  analysis,  without  first  examining  our  assumed  law  of  optics.  That  law 
should  undoubtedly  read,  “ the  intensity  of  a light  at  any  [sensible]  distance  is 
equal  to  its  intensity  at  the  distance  unity,  divided  by  the  square  of  that  dis- 
tance.” We  should  then  be  saved  from  the  hypothesis  in  question,  and  from 
the  physical  anomaly. 

But  in  further  support  of  the  analysis,  and  to  show  how  wondrously  it  re- 
sponds to  every  hypothesis,  let  us  rid  the  question  entirely  of  physical  phenom- 
ena, and  present  it  as  an  abstract  question. 

Let  y represent  the  general  ordinate  of  a curve  which  shall  be  equal  to  a 
when  the  abscissa  is  made  equal  to  unity,  and  which  shall  vary  inversely  as  the 
square  of  the  abscissa.  Then, 


will  be  the  equation  of  such  curve.  This  equation  may  be  readily  constructed, 
and  we  shall  have  a curve  with  two  branches,  BC  and  de  as  shown  in  Fig.  1. 


By  making  x and  y zero  in  succession  in  equation  (1),  it  appears  at  once  that  the 
axes  are  asymptotes  to  either  branch,  and  thus  that  y is  infinite  when  * is  zero, 
and  zero  when  x is  infinite. 

Let  us  now  have  auother  such  curve  whose  equation  is, 


b 

(c  — X)2 


(2). 


This  curve  gives  y equal  to  b,  when  c—x  is  equal  to  unity ; and  the  ordinate 
whose  abscissa  is  equal  to  c,  is  an  asymptote  to  the  curve. 

Now,  let  it  be  required  to  find  the  common  points  of  these  two  curves  ; that 
is,  the  points  whose  ordinates  (they  corresponding  to  the  intensities  in  the  prob- 
lem) are  equal.  From  (1)  and  (2),  we  must  have, 


a _ b 
x 2 (c—x)2  ’ 

the  old  equation,  but  with  no  question  of  light  in  it.  The  values  of  x may  now 
be  found  as  before,  and  the  points  constructed.  The  curves  would,  under  the 
first  hypothesis  in  the  problem,  take  the  general  form  shown  in  Fig.  2. 


156  ELEMENTS  OP  ALGEBRA. 


These  curves  will  show  to  the  eye  the  results  of  all  the  hypotheses  which 
may  he  made  in  the  problem.  We  shall  make  only  the  one  in  dispute,  namely, 
c=0,  and  a>  or  <b.  The  hypothesis  c=0  makes  the  asymptotical  ordinate  co- 
incide with  the  axis  of  Y,  and  the  curves  will  assume  the  position  shown  in 
Fig.  3. 


It  will  be  seen  that  both  curves  approach  the  axis  of  Y,  and  that  neither  can 
ever  reach  it — the  ordinate  at  the  point  zero  being  infinite — that  is,  resuming 
the  question  of  lights,  the  intensities  of  both  lights  are  infinite  at  the  origin, 
and  that  point  is  equally  illuminated.  It  will  hardly  be  said  that  one  of  these 
curves  reaches  the  axis  before  the  other. 


THE  END. 


Date  Due 


L.  B.  Cat.  No.  1 137 


512.03  S559E 

StlQUT) 


5518 


-Elements  of  Alggh-ra 


512.02  S559E 


5518 


